2.
o
Quadratic Equations
Introduction : :
An equation involving one variable with highest power of variable as 2 is
called a quadratic equation.
Example : x2 + 6x + 7 = 0, 2x2 + 7 = 0
The general form of quadratic equation is ax2 + bx + c = 0 where a, b and c
are real numbers and a  0.
Things To Remember :
•
•
•
Highest power of variable has to be 2.
Equation has to be in simplified form.
Simplification implies having all the terms involving variable in the
numerator. If variables are in the denominator, then each term of equation
has to be multiplied by the term of denominator so that variable in the
denominator gets cancelled.
EXERCISE - 2.1 (TEXT BOOK PAGE NO. 36) :
1.
(i)
Sol.
(ii)
Sol.
(iii)
Sol.
Which of the following are quadratic equations ?
11 = – 4x2 – x3
11 = 4x2 – x3
 x3 + 4x2 + 11 = 0
Here maximum index of the variable x is 3.
So it is not a quadratic equation.
3 2
y = 2y + 7
4
3
– y2 = 2y + 7
4
3 2
 –
y – 2y – 7 = 0
4
3
Here a = – , b = –2, c = –7 are real numbers
4
where a  0
So it is a quadratic equation in variable y.
–
(1 mark)
(1 mark)
(y – 2) (y + 2) = 0
(1 mark)
(y – 2) (y + 2) = 0
 (y)2 – (2)2 = 0
[ a2 – b2 = (a + b) (a – b)]
2
 y –4=0
 y2 + 0y – 4 = 0
Here a = 1, b = 0, c = – 4 are real numbers
where a  0
So it is a quadratic equation in variable y.
S C H O O L S E C TI O N
49
MT
ALGEBRA
(iv)
Sol.
3
–4=y
y
EDUCARE LTD.
(1 mark)
3
–4=y
y
Multiplying throughout by y, we get,
3 – 4y
= y2
 – y2 – 4y + 3
= 0
Here a = –1, b = – 4, c = 3 are real numbers
where a  0
So it is a quadratic equation in variable y.
(v)
Sol.
m3 + m + 2 = 4m
m3 + m + 2 = 4m

m3 + m – 4m + 2
= 0

m3 – 3m + 2
= 0
Here the maximum index of the variable m is 3.
So it is not a quadratic equation.
(1 mark)
(vi)
Sol.
n – 3 = 4n
n – 3 = 4n

n – 4n – 3
= 0

– 3n – 3
= 0
 0n2 – 3n – 3
= 0
Here a = 0
So it is not a quadratic equation.
(1 mark)
(vii)
Sol.
y2 – 4 = 11y
y2 – 4 = 11y
 y2 – 11y – 4
= 0
Here a = 1, b = –11, c = – 4 are real numbers
where a  0
So it is a quadratic equation in variable y.
(1 mark)
(viii)
Sol.
(ix)
Sol.
7
= 4z + 5
z
7
z–
= 4z + 5
z
Multiplying throughout by z, we get,
z2 – 7
= 4z2 + 5z
2
2
 z – 4z – 5z – 7 = 0
 –3z2 – 5z – 7
= 0
Here a = –3, b = –5, c = –7 are real numbers
where a  0
So it is a quadratic equation in variable z.
z–
3y2 – 7 =
 3y2 –
(1 mark)
3y
3y2 – 7 =
(1 mark)
3y
3 y – 7 =0
Here a = 3, b = – 3 , c = –7 are real numbers
where a  0
So it is a quadratic equation in variable y.
50
S C H O O L S E C TI O N
MT
(x)
EDUCARE LTD.
q2 – 4
= –3
q2
ALGEBRA
(1 mark)
q2 – 4
= –3
q2
Sol.




q2 – 4
= –3q2
q2 + 3q2 – 4
= 0
2
4q – 4
= 0
4q2 + 0q – 4
= 0
Here a = 4, b = 0, c = – 4 are real numbers
where a  0
So it is a quadratic equation in variable q.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :
4.
(i)
Sol.
Which of the following equations are quadratic ?
13 = – 5y2 – y3
13 = – 5y2 – y3
 y3 + 5y2 + 13
= 0
Here maximum index of the variable y is 3.
So it is not a quadratic equation.
(ii)
–
Sol.
(iii)
Sol.
(iv)
Sol.
5 2
x = 2x + 9
3
(1 mark)
(1 mark)
5 2
x = 2x + 9
3
5
 – x2 – 2x – 9 = 0
3
5
Here a = – , b = – 2, c = – 9 are real numbers.
3
Where a  0.
So it is a quadratic equation in variable x.
–
(x + 3) (x – 4) = 0
(x + 3) (x – 4) = 0
x (x – 4) + 3 (x – 4)
= 0

x2 – 4x + 3x – 12
= 0

x2 – x – 12
= 0
Here a = 1, b = – 1, c = – 12 are real numbers.
Where a  0.
So it is a quadratic equation in variable x.
5
– 3 = x2
x
(1 mark)
(1 mark)
5
– 3 = x2
x
Multiplying throughout by x, we get;
5 – 3x
= x3

0
= x3 + 3x – 5

x3 + 3x – 5
= 0
Here maximum index of the variable x is 3.
So it is not a quadratic equation.
S C H O O L S E C TI O N
51
MT
ALGEBRA
EDUCARE LTD.
(v)
Sol.
n3 – n + 4 = n 3
n3 – n + 4 = n3
 n3 – n3 – n + 4 = 0

–n+4
= 0
Here a = 0
So it is not a quadratic equation.
(1 mark)
(vi)
Sol.
n – 3 = 4n2
n – 3 = 4n2
 – 4n2 + n – 3
= 0
Here a = – 4, b = 1, c = – 3 are real numbers.
Where a  0
So it is a quadratic equation in variable n.
(1 mark)
(vii)
Sol.
x2 + 4x = 11
x2 + 4x = 11
 x3 + 4x – 11
= 0
Here a = 1, b = 4, c = – 11 are real numbers.
Where a  0
So it is a quadratic equation in variable x.
(1 mark)
(viii)
Sol.
o
5
= 4m + 5
m
5
m–
= 4m + 5
m
Multiplying throughout by m, we get;
m2 – 5
= 4m2 + 5m

0
= 4m2 – m2 + 5m + 5
2
 3m + 5m + 5 = 0
Here a = 3, b = 5, c = 5 are real numbers.
Where a  0
So it is a quadratic equation in variable m.
(1 mark)
m–
Standard form of a quadratic equations :
The standard form of a quadratic equation in variable x is ax2 + bx + c = 0
where a, b and c are real numbers and a  0.
EXERCISE - 2.1 (TEXT BOOK PAGE NO. 36) :
2.
(i)
Sol.
(ii)
Sol.
(iii)
Sol.
52
Write the following quadratic equations in standard form
ax2 + bx + c = 0
7 – 4x –x2 = 0
7 – 4x – x2 = 0
 – x2 – 4x + 7
= 0
Multiplying throughout by –1, we get,
x2 + 4x – 7
= 0
3y2 = 10y + 7
3y2 = 10y + 7
 3y2 – 10y – 7
(1 mark)
(1 mark)
=
0
(m + 4) (m – 10) = 0
(m + 4) (m – 10) = 0
 m(m – 10) + 4(m – 10)

m2 – 10m + 4m – 40

m2 – 6m – 40
(1 mark)
=
=
=
0
0
0
S C H O O L S E C TI O N
MT
(iv)
Sol.
(v)
Sol.
(vi)
Sol.
(vii)
Sol.
(viii)
Sol.
(ix)
Sol.
p(p – 6) = 0
p(p – 6) = 0

p2 – 6p
2

p – 6p + 0
Sol.
(1 mark)
=
=
0
0
x2
–4=0
25
x2
–4=0
25
Multiplying throughout by 25, we get,
x2 – 100
= 0
2
 x + 0x – 100
= 0
7
=4
n
7
n–
=4
n
Multiplying throughout by n, we get,
n2 – 7
= 4n
2

n – 4n – 7
= 0
(1 mark)
n–
(1 mark)
y2 – 9 = 13y
y2 – 9 = 13y
 y2 – 13y – 9
(1 mark)
=
0
5
=z–6
z
5
2z –
=z–6
z
Multiplying throughout by z, we get,
2z2 – 5
=
z2 – 6z
 2z2 – z2 + 6z – 5 = 0

z2 + 6z – 5
= 0
2z –
x2 = –7 –
(1 mark)
(1 mark)
10 x
x2 = –7 –
 x2 +
(x)
ALGEBRA
EDUCARE LTD.
10 x
10 x + 7
m2 + 5
= –3
m2
m2  5
= –3
m2

m2 + 5
2
 m + 3m2 + 5

4m2 + 5
2
 4m + 0m+ 5
=
0
(1 mark)
=
=
=
=
–3m2
0
0
0
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :
6.
(i)
Sol.
Write the quadratic equations in standard from ax2 + bx + c = 0
8 – 3x – 4x2 = 0
(1 mark)
8 – 3x – 4x2 = 0
0
= 4x2 + 3x – 8
2
 4x + 3x – 8
= 0
S C H O O L S E C TI O N
53
MT
ALGEBRA
(ii)
Sol.
3y2 = 8y – 5
3y2 = 8y – 5
 3y2 – 8y + 5
EDUCARE LTD.
(1 mark)
=
0
(iii)
Sol.
(x + 5) (x – 11) = 0
(x + 5) (x – 11) = 0
 x2 – 11x + 5x – 55 = 0

x2 – 6x – 55 = 0
(1 mark)
(iv)
Sol.
m (m – 7) = 0
m (m – 7) = 0

m2 – 7m
2
 m – 7m + 0
(1 mark)
(v)
Sol.
(vi)
Sol.
(vii)
Sol.
(viii)
Sol.
=
=
0
0
y2
–3=0
23
(1 mark)
y2
–3=0
23
Multiplying throughout by 23, we get
y2 – 69
= 0
2
 y + 0y – 69
= 0
x–
6
=5
x
(1 mark)
6
=5
x
Multiplying throughout by x, we get
x2 – 6
= 5x
2

x – 5x – 6
= 0
x–
– x2 – 5 = 16x
– x2 – 5 = 16x
 0 = x2 + 16x + 5
 x2 + 16x + 5
=
(1 mark)
0
5
y – y = 3y – 7
(1 mark)
5
y – y = 3y – 7
Multiplying throughout by y, we get
y2 – 5 = 3y2 – 7y

0 = 3y2 – y2 – 7y + 5

0 = 2y2 – 7y + 5
2
 2y – 7y + 5 = 0
(ix)
Sol.
y2 = 5 –
 y2 +
54
(1 mark)
7y
y =5–
2
7y
7y–5
=
0
S C H O O L S E C TI O N
MT
(x)
x2 – 7
=7
x
Sol.


(xi)
(1 mark)
x2 – 7
=7
x
x2 – 7
2
x – 7x – 7
4z2 + 5 =
Sol.
=
=
7x
0
(1 mark)
3z
4z2 + 5 =
 4z2 –
(xii)
ALGEBRA
EDUCARE LTD.
3z
3z + 5
=
0
p2 + 5
=–7
3p2
Sol.




p2 + 5
=–7
3p2
p2 + 5
p2 + 5
2
p + 21p2 + 5
22p2 + 0p + 5
(1 mark)
=
=
=
=
– 7 × 3p2
– 21p2
0
0
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :
1.
(i)
Sol.
Find the values of a, b, c for following quadratic equations by comparing
with standard form :
x2 + 2x + 1 = 0
(1 mark)
x2 + 2x + 1 = 0
Comparing with ax2 + bx + c = 0
a = 1, b = 2, c = 1
(ii)
Sol.
2x2 – x + 3 = 0
2x2 – x + 3 = 0
Comparing with ax2 + bx + c = 0
a = 2, b = – 1 c = 3
(1 mark)
(iii)
Sol.
x2 – x – 3 = 0
x2 – x – 3 = 0
Comparing with ax2 + bx + c = 0
a = 1, b = – 1, c = – 3
(1 mark)
(iv)
Sol.
x2 + 5x – 4 = 0
x2 + 5x – 4 = 0
Comparing with ax2 + bx + c = 0
a = 1, b = 5, c = – 4
(1 mark)
(v)
Sol.
x2 – 7x + 4 = 0
x2 – 7x + 4 = 0
Comparing with ax2 + bx + c = 0
a = 1, b = – 7, c = 4
(1 mark)
S C H O O L S E C TI O N
55
ALGEBRA
(vi)
Sol.
o
4x2 – 9x + 1 = 0
4x2 – 9x + 1 = 0
Comparing with ax2 + bx + c = 0
a = 4, b = – 9, c = 1
MT
EDUCARE LTD.
(1 mark)
Roots/solution of a quadratic equation :
Any value of the variable which when substituted in a given quadratic
equation makes the left hand side of the equation equal to its right hand
side, is said to be a solution of the equation.
A solution of a quadratic equation is also called a root of that equation.
The set of roots of a quadratic equation is called the solution set of the
quadratic equation.
For example : Consider x2 – 4x + 3 = 0 take sum real numbers as x = 1, 2
and 3 and see whether the equation is satisfied or not.
(i) L.H.S. = 12 – 4 (1) + 3 = 1 – 4 + 3 = –3 + 3 = 0
 L.H.S. = R.H.S.
 x = 1 is the root of the given quardratic equation
(ii) L.H.S. = (2)² – 4 (2) + 3 = 4 – 8 + 3 = – 4 + 3 = –1
 L.H.S.  R.H.S.
 x = 2 is not the root of the given quardratic equation
(iii) L.H.S. = (3)² – 4 (3) + 3 = 9 – 12 + 3 = 12 – 12 = 0
 L.H.S. = R.H.S.
 x = 3 is the root of the given quardratic equation
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :
3.
(i)
Sol.
Determine whether the given values of ‘x’ is a roots of given quadratic equation.
(1 mark)
x2 – 2x + 1 = 0, x = 1
2
x – 2x + 1 = 0, x = 1
Putting x = 1 in L.H.S. we get,
L.H.S. = (1)2 – 2 (1) + 1
= 1 – 2 (1) + 1
= 2–2
= 0
= R.H.S.
 L.H.S. = R.H.S.
Thus equation is satisfied.
So 1 is the root of the given quadratic equation.
(ii)
Sol.
x2 + 2x + 1 = 0, x = – 1
x2 + 2x + 1 = 0, x = – 1
Putting x = – 1 in L.H.S. we get,
L.H.S. = (– 1)2 + 2 (– 1) + 1
= 1–2+1
= 2–2
= 0
= R.H.S.
 L.H.S. = R.H.S.
Thus equation is satisfied.
So – 1 is the root of the given quadratic equation.
56
(1 mark)
S C H O O L S E C TI O N
MT
EDUCARE LTD.
ALGEBRA
(iii)
Sol.
x 2 – x = 0, x = 0
x2 – x = 0, x = 0
Putting x = 0 in L.H.S., we get,
L.H.S. = (0)2 – 0
= 0–0
= 0
= R.H.S.
 L.H.S. = R.H.S.
Thus equation is satisfied.
So 0 is the root of the given quadratic equation.
(1 mark)
(iv)
Sol.
x2 + x – 1 = 0, x = 2
x2 + x – 1 = 0, x = 2
Putting x = 2 in L.H.S., we get,
L.H.S. = (2)2 + 2 – 1
= 4+1
= 5
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So 2 is not the root of the given quadratic equation.
(1 mark)
(v)
Sol.
x2 – 4x + 4 = 0, x = 0
x2 – 4x + 4 = 0, x = 0
Putting x = 0 in L.H.S., we get,
L.H.S. = (0)2 + 4 (0) + 4
= 0–0+4
= 4
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So 0 is not the root of the given quadratic equation.
(1 mark)
(vi)
Sol.
x2 – 4x + 1 = 0, x = 1
x2 – 4x + 1 = 0, x = 1
Putting x = 1 in L.H.S., we get,
L.H.S. = (1)2 – 4 (1) + 1
= 1–4+1
= 2–4
= –2
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So 1 is not the root of the given quadratic equation.
(1 mark)
EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :
1.
(i)
Sol.
In each of the examples given below determine whether the values
given against each of the quadratic equation are the roots of the equation
or not.
x2 + 3x – 4 = 0, x = 1, –2, – 3
(3 marks)
a) By putting x = 1 in L.H.S. we get
L.H.S.
=
(1)2 + 3(1) – 4
=
1+3–4
=
4–4
=
0
S C H O O L S E C TI O N
57
MT
ALGEBRA
EDUCARE LTD.
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
So 1 is the root of the given quadratic equation.
b) By putting x = –2 in L.H.S. we get
L.H.S.
=
(–2)2 + 3(–2) – 4
=
4–6–4
=
–6

R.H.S.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
So –2 is not the root of the given quadratic equation.
c) By putting x = –3 in L.H.S. we get
L.H.S.
=
(–3)2 + 3(–3) – 4
=
9–9–4
=
–4

R.H.S.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
So –3 is not the root of the given quadratic equation.
(ii)
Sol.
2 3
,
3 2
a) By putting m = 2 in L.H.S. we get
L.H.S.
=
4(2)2 – 9
=
4(4) – 9
=
16 – 9
=
7

R.H.S.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
So 2 is not the root of the given quadratic equation.
4m2 – 9 = 0, m = 2,
b) By putting m =
L.H.S.
=
=
(3 marks)
2
in L.H.S. we get
3
2
 2
4  – 9
 3
4
–9
4×
9
16
–9
9
16 – 81
=
9
– 65
=
9

R.H.S.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
=
So
58
2
is not the root of the given quadratic equation.
3
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
3
in L.H.S. we get
2
2
 3
L.H.S.
=
4  – 9
 2
9
=
4×
–9
4
=
9–9
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
3
So
is the root of the given quadratic equation.
2
c) By putting m =
(iii)
Sol.
x2 + 5x – 14 = 0, x =
2 , –7, 3
a) By putting x = 2 in L.H.S. we get
L.H.S.
=
( 2 )2 + 5( 2 ) – 14
(3 marks)
2 + 5 2 – 14
=
5 2 – 12

R.H.S.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
So 2 is not the root of the given quadratic equation.
=
b) By putting x = –7 in L.H.S. we get
L.H.S.
=
(–7)2 + 5(–7) – 14
=
49 – 35 – 14
=
49 – 49
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
So –7 is the root of the given quadratic equation.
c) By putting x = 3 in L.H.S. we get
L.H.S.
=
(3)2 + 5(3) – 14
=
9 + 15 – 14
=
24 – 14
=
10

R.H.S.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
So 3 is not the root of the given quadratic equation.
(iv)
Sol.
1
, –3
2
a) By putting p = 1 in L.H.S. we get
L.H.S.
=
2(1)2 + 5(1) – 3
=
2(1) + 5 – 3
=
2+5–3
=
7–3
=
4

R.H.S.
2p2 + 5p – 3 = 0, p = 1,
S C H O O L S E C TI O N
(3 marks)
59
MT
ALGEBRA
EDUCARE LTD.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
So 1 is not the root of the given quadratic equation.
b) By putting p =
L.H.S.
=
1
in L.H.S. we get
2
2
 1
 1
2   + 5   – 3
 2
2
1
1
+5×
–3
4
2
=
2×
=
1
5
+
–3
2
2
6
–3
2
=
3–3
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
=
So
1
is the root of the given quadratic equation.
2
c) By putting p = – 3 in L.H.S. we get
L.H.S.
=
2(– 3)2 + 5(– 3) – 3
=
2(9) – 15 – 3
=
18 – 18
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
So – 3 is the root of the given quadratic equation.
(v)
Sol.
60
n2 + 4n = 0, n = 0, – 2, – 4
a) By putting n = 0 in L.H.S. we get
L.H.S.
=
(0)2 + 4(0)
=
0+0
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
So 0 is the root of the given quadratic equation.
b) By putting n = –2 in L.H.S. we get
L.H.S.
=
(– 2)2 + 4(– 2)
=
4–8
=
–4

R.H.S.
 L.H.S.

R.H.S.
Thus equation is not satisfied.
So –2 is not the root of the given quadratic equation.
(3 marks)
S C H O O L S E C TI O N
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c) By putting n = –4 in L.H.S. we get
L.H.S.
=
(– 4)2 + 4(– 4)
=
16 – 16
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
So – 4 is the root of the given quadratic equation.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
10.
(i)
Sol.
In each of the examples given below determine whether the values
given against each of the quadratic equation are the roots of the equation
or not.
y2 + 5y + 6 = 0;
y = 4, – 2, – 3
(3 marks)
y2 + 5y + 6 = 0; y = 4, – 2, – 3
(a) By putting y = 4 in L.H.S. we get
L.H.S. = (4)2 + 5 (4) + 6
= 16 + 20 + 6
= 42
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So 4 is not the root of the given quadratic equation.
(b) By putting y = – 2 in L.H.S. we get
L.H.S. = (– 2)2 + 5 (– 2) + 6
= 4 – 10 + 6
= 10 – 10
= 0
= R.H.S.
 L.H.S.  R.H.S.
Thus equation is satisfied.
So – 2 is the root of the given quadratic equation.
(c) By putting y = – 3 in L.H.S. we get
L.H.S. = (– 3)2 + 5 (– 3) + 6
= 9 – 15 + 6
= 15 – 15
= 0
= R.H.S.
 L.H.S.  R.H.S.
Thus equation is satisfied.
So – 3 is the root of the given quadratic equation.
(ii)
Sol.
25x2 – 36 = 0; x = 2,
6 –6
,
5
5
(3 marks)
6 –6
,
5
5
(a) By putting x = 2 in L.H.S. we get
L.H.S. = 25(2)2 – 36
= 25 (4) – 36
= 100 – 36
= 64
25x2 – 36 = 0;
S C H O O L S E C TI O N
x = 2,
61
MT
ALGEBRA
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 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So 2 is not the root of the given quadratic equation.
(b) By putting x =
L.H.S. =
6
in L.H.S. we get
5
 6
25  
 5
2
– 36
36
– 36
25
= 36 – 36
= 0
= R.H.S.
 L.H.S.  R.H.S.
Thus equation is satisfied.
=
So
25 ×
6
is the root of the given quadratic equation.
5
(c) By putting x =
L.H.S. =
6
in L.H.S. we get
5
  6
25 
 5 
2
– 36
36
– 36
25
= 36 – 36
= 0
= R.H.S.
 L.H.S.  R.H.S.
Thus equation is satisfied.
=
So
(iii)
Sol.
25 ×
6
is the root of the given quadratic equation.
5
x2 – 7x – 18 = 0;
x = 3 , – 2, 4
(3 marks)
x2 – 7x – 18 = 0; x = 3 , – 2, 4
(a) By putting x =
L.H.S. =
=
3 in L.H.S. we get
( 3 )2 – 7 ( 3 ) – 18
3 – 7 3 – 18
=
– 15 – 7 3
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So
62
3 is not the root of the given quadratic equation.
S C H O O L S E C TI O N
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(b) By putting x = – 2 in L.H.S. we get
L.H.S. = (– 2)2 – 7 (– 2) – 18
= 4 + 14 – 18
= 18 – 18
= 0
= R.H.S.
 L.H.S.  R.H.S.
Thus equation is satisfied.
So – 2 is the root of the given quadratic equation.
(c) By putting x = 4 in L.H.S. we get
L.H.S. = (4)2 – 7 (4) – 18
= 16 – 28 – 18
= 16 – 46
= – 30
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So 4 is not the root of the given quadratic equation.
(iv)
Sol.
1
,3
2
1
3x2 – 8x – 3 = 0;x = 9, , 3
2
(a) By putting x = 9 in L.H.S. we get
L.H.S. = 3 (9)2 – 8 (9) – 3
= 3 (81) – 72 – 3
= 243 – 75
= 168
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So 9 is not the root of the given quadratic equation.
3x2 – 8x – 3 = 0;
x = 9,
(b) By putting x =
1
in L.H.S. we get
2
L.H.S. =
 1
3  
 2
2
(3 marks)
 1
– 8   – 3
2
1
8
–
–3
4
2
=
3×
=
3
–4–3
4
3
–7
4
3  28
=
4
 25
=
4
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
=
So
1
is not the root of the given quadratic equation.
2
S C H O O L S E C TI O N
63
ALGEBRA
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(c) By putting x = 3 in L.H.S. we get
L.H.S. = 3 (3)2 – 8 (3) – 3
= 3 (9) – 24 – 3
= 27 – 27
= 0
= R.H.S.
 L.H.S. = R.H.S.
Thus equation is satisfied.
So 3 is the root of the given quadratic equation.
(v)
Sol.
x2 – 4x = 0; x = 0, – 2, 4
x2 – 4x = 0; x = 0, – 2, 4
(a) By putting x = 0 in L.H.S. we get
L.H.S. = (0)2 – 4 (0)
= 0–0
= 0
= R.H.S.
 L.H.S. = R.H.S.
Thus equation is satisfied.
So 0 is the root of the given quadratic equation.
(3 marks)
(B) By putting x = – 2 in L.H.S. we get
L.H.S. = (– 2)2 – 4 (– 2)
= 4+8
= 12
 R.H.S.
 L.H.S.  R.H.S.
Thus equation is not satisfied.
So – 2 is not the root of the given quadratic equation.
(C) By putting x = 4 in L.H.S. we get
L.H.S. = (4)2 – 4 (4)
= 16 – 16
= 0
= R.H.S.
 L.H.S. = R.H.S.
Thus equation is satisfied.
So 4 is the root of the given quadratic equation.
EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :
4.
Sol.
64
State whether k is the root of the given equation y2 – (k – 4)y – 4k = 0.
(4 marks)
2
y – (k – 4)y – 4k = 0.
By putting y = k in L.H.S. we get
L.H.S.
=
(k)2 – (k – 4)(k) – 4k
=
k2 – (k2 – 4k) – 4k
=
k2 – k2 + 4k – 4k
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
So k is the root of the given quadratic equation.
S C H O O L S E C TI O N
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PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
11.
–k
is the root of the quadratic equation
2
2x2 + (k – 6) x – 3k = 0
State whether x =
Sol.
2x2 + (k – 6) x – 3k = 0; x =
(3 marks)
–k
2
–k
in L.H.S. we get
2
Putting x =
2
L.H.S.
=
 –k 
 –k
2
  (k – 6) 
 – 3k
 2 
 2 
=
2
=
k 2 (–k 2  6k)

– 3k
2
2
=
k 2 – k 2  6k
– 3k
2
k 2 (k 2  6k)

– 3k
4
2
6k
– 3k
2
=
3k – 3k
=
0
=
R.H.S.
 L.H.S.
=
R.H.S.
Thus equation is satisfied.
=
So
–k
is the root of the given quadratic equation.
2
EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :
2.
Sol.
If one root of the quadratic equation x2 – 7x + k = 0 is 4, then find the
value of k.
(2 marks)
x2 – 7x + k = 0
x = 4 is the root of given quadratic equation.
So it satisfies the given equation.
 (4)2 – 7(4) + k = 0

16 – 28 + k
= 0

–12 + k
= 0

k
=
12
3.
If one root of the quadratic equation 3y2 – ky + 8 = 0 is
Sol.
value of k.
3y2 – ky + 8 = 0
2
, then find the
3
(3 marks)
2
is the root of given quadratic equation.
3
So it satisfies the given equation.
y =
 2
 3 
 3
2
S C H O O L S E C TI O N
 2
– k   + 8
3
= 0
65
MT
ALGEBRA

34
2k
–
+8
9
3
= 0
4
2k
–
+8
=
3
3
Multiplying throughout
4 – 2k + 24
=

– 2k
=


k

k
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0
by 3, we get,
0
–28
28
2
= 14
=
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :
7.
Sol.
Find the value of k if x = 4 is the solution of the equation
3x2 + kx – 2 = 0
(3 marks)
2
3x + kx – 2 = 0
x = 4 is the solution of given quadratic equation.
Substituting x = 4 in given quadratic equation, it will get satisfied.

3 (4)2 + k (4) – 2
= 0
2

3 (16) + 4k – 2
= 0

48 + 4k – 2
= 0

4k + 46
= 0

4k
= – 46

k
=

k
=
 46
4
 23
2
EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :
5.
Sol.
If one root of the quadratic equation kx2 – 7x + 12 = 0 is 3, then find the
(2 marks)
value of k.
2
kx – 7x + 12 = 0
x = 3 is root of given quadratic equation.
So it satisfies the given equation

k(3)2 – 7(3) + 12
= 0

k(9) – 21 + 12
= 0

9k – 9
= 0

9k
= 9
9

k
=
9

o
(I)
(II)
(III)
66
k
= 1
Method for solving quadratic equations :
Factorisation method
Completing square method
Formula method
S C H O O L S E C TI O N
MT
o
ALGEBRA
EDUCARE LTD.
Method 1 : FACTORIZATION
In this method, we solve quadratic equations by finding the factors of it.
Factors can be found either by splitting the middle term, using the formula
a2 – b2 = (a + b) (a – b) or taking the common terms and writing the factors
in product form.
If the given equation ax2 + bx + c = 0 can be written as
(px + q) (mx + n)
= 0

(px + q) = 0
or (mx + n) = 0

px = – q
or mx = – n
–q
x= p


or
x=
–n
m
–q
–n
p and m are the roots of the given quadratic equation.
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :
1.
(i)
Sol.
Solve the following quadratic equations by factorization method.
x2 – 5x + 6 = 0
(2 marks)
x2 – 5x + 6 = 0

x2 – 3x – 2x + 6 = 0
 x(x – 3) – 2(x – 3) = 0

(x – 3) (x – 2) = 0

x – 3 = 0 or x – 2 = 0

x = 3 or x = 2
(ii)
Sol.
x2 + 10x + 24 = 0
x2 + 10x + 24 = 0

x2 + 6x + 4x + 24
 x(x + 6) + 4(x + 6)

(x + 6) (x + 4)

x+6=0
=
=
=
or

or x = –4
(iii)
Sol.
x = –6
x2 – 13x – 30 = 0
x2 – 13x – 30 = 0
 x2 – 15x + 2x – 30
 x (x – 15) + 2 (x– 15)

(x – 15) (x + 2)

x – 15 = 0

x = 15
(2 marks)
0
0
0
x+4=0
(2 marks)
=
=
=
or
or
0
0
0
x+2=0
x = –2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
9.
(iii)
Sol.
Solve the following equations by factorization method.
y2 + 8y + 16 = 0
y2 + 8y + 16 = 0

y2 + 4y + 4y + 16 = 0
 y (y + 4) + 4 (y + 4) = 0

(y + 4) (y + 4) = 0

(y + 4)2 = 0
Taking square root on both the sides, we get,
y+4 = 0

y = –4

S C H O O L S E C TI O N
y
=
(2 marks)
–4
67
MT
ALGEBRA
(iv)
Sol.
y2 + 3y
y2 +

y2
 y (y



(v)
Sol.
(vi)
Sol.
(ix)
Sol.
– 18 = 0
3y – 18 = 0
– 3y + 6y – 18
– 3) + 6 (y – 3)
(y – 3) (y + 6)
y–3=0
y=3
(2 marks)
=
=
=
or
0
0
0
y+6=0
or y = – 6
(2 marks)
x2 + 5x + 6 = 0
x2 + 5x + 6 = 0

x2 + 2x + 3x + 6
 x (x + 2) + 3 (x + 2)

(x + 2) (x + 3)

x+2=0
=
=
=
or

or x = – 3
x=–2
0
0
0
x+3=0
(2 marks)
y2 – 16y + 63 = 0
y2 – 16y + 63 = 0

y2 – 9y – 7y + 63
 y (y – 9) – 7 (y – 9)

(y – 9) (y – 7)

y–9=0
=
=
=
or

or y = 7
y=9
0
0
0
y–7=0
(2 marks)
y2 – 5y – 24 = 0
y2 – 5y – 24 = 0

y2 – 8y + 3y – 24
 y (y – 8) + 3 (y – 8)

(y – 8) (y + 3)

y–8=0
=
=
=
or

or y = – 3
y=8
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0
0
0
y+3=0
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :
1.
(iv)
Sol.
Solve the following quadratic equations by factorization method.
(2 marks)
x2 – 17x + 60 = 0
x2 – 17x + 60 = 0

x2 – 12x – 5x + 60 = 0
 x(x – 12) – 5(x – 12) = 0

(x – 12) (x – 5) = 0

x – 12 = 0 or x – 5 = 0

(vii)
Sol.
68
x = 12 or x = 5
(2 marks)
x2 = 2(11x – 48)
x2 = 2(11x – 48)

x2
2

x – 22x + 96

x2 – 16x – 6x + 96
 x (x – 16) – 6 (x – 16)

(x – 16) (x – 6)

x – 16 = 0
=
=
=
=
=
or

or x = 6
x = 16
22x – 96
0
0
0
0
x–6=0
S C H O O L S E C TI O N
MT
(viii)
Sol.
(x)
Sol.
ALGEBRA
EDUCARE LTD.
(2 marks)
21x = 196 – x2
21x = 196 – x2

x2 + 21x – 196
2

x – 28x – 7x – 196

x (x + 28) – 7 (x + 28)

(x + 28) (x – 7)

x + 28 = 0
=
=
=
=
or
0
0
0
0
x–7=0

or
x=7
x = –28
(2 marks)
x2 – x – 132 = 0
x2 – x – 132 = 0

x2 – 12x + 11x – 132
 x (x – 12) + 11 (x – 12)

(x – 12) (x + 11)

x – 12 = 0
=
=
=
or
0
0
0
x + 11 = 0

or
x = –11
x = 12
(xxiii) x2 – 3 3 x + 6 = 0
Sol.
x2 – 3 3 x + 6 = 0

x2 – 2 3 x – 3 x + 6
2

x – 2 3x – 3x + 2 × 3
 x(x – 2 3 ) – 3 (x – 2 3 )

(x – 2 3 ) (x – 3 )

x– 2 3 =0

x= 2 3
(3 marks)
=
=
=
=
or
0
0
0
0
x–
or x =
3 =0
3
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
18.
(ii)
Sol.
Solve the following quardratic equations by factorization method.
y2 – 3 = 0
(2 marks)
2
y –3=0


(y +


(y)2 – ( 3 )2
=
0
3 ) (y –
=
0
y+
3)
3 =0
y=–
3
or y –
3 =0
or y =
3
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :
1.
(xi)
Sol.
Solve the following quadratic equations by factorization method.
(3 marks)
5x2 – 22x – 15 = 0
5x2 – 22x – 15 = 0

5x2 – 25x + 3x – 15 = 0

5x(x – 5) + 3(x – 5) = 0

(x – 5) (5x + 3) = 0

x – 5 = 0 or 5x + 3 = 0

x = 5 or 5x = –3

S C H O O L S E C TI O N
x=5
or x =
–3
5
69
MT
ALGEBRA
(xii)
Sol.
(xiii)
Sol.
(xiv)
Sol.
(xv)
Sol.
(xvi)
Sol.
70
(3 marks)
3x2 – x – 10 = 0
3x2 – x – 10 = 0
 3x2 – 6x + 5x – 10
 3x(x – 2) + 5(x – 2)

(x – 2) (3x + 5)

x–2=0

x=2
=
=
=
or
or

or x =
x=2
0
0
0
3x + 5 = 0
3x = –5
–5
3
(3 marks)
2x2 – 5x – 3 = 0
2x2 – 5x – 3 = 0

2x2 – 6x + x – 3
 x(x – 3) + 1(x – 3)

(x – 3) (2x + 1)

x–3=0

x=3
=
=
=
or
or

or x =
x=3
0
0
0
2x + 1 = 0
2x = –1
–1
2
(3 marks)
x (2x + 3) = 35
x (2x + 3) = 35

2x2 + 3x – 35
2

2x + 10x – 7x – 35
 2x (x + 5) – 7 (x + 5)

(x + 5) (2x – 7)

x+5=0

x=–5
=
=
=
=
or
or

or x =
x=–5
0
0
0
0
2x – 7 = 0
2x = 7
7
2
(3 marks)
7x2 + 4x – 20 = 0
7x2 + 4x – 20 = 0

7x2 + 14x – 10x – 20
 7x (x + 2) – 10 (x + 2)

(x + 2) (7x – 10)

x+2=0

x=–2
=
=
=
or
or

or x =
x=–2
10x2 + 3x – 4 = 0
10x2 + 3x – 4 = 0

10x2 + 8x – 5x – 4
 2x (5x + 4) – 1 (5x + 4)

(5x + 4) (2x – 1)

5x + 4 = 0

5x = – 4

x=
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–4
5
0
0
0
7x – 10 = 0
7x = 10
10
7
(3 marks)
=
=
=
or
or
0
0
0
2x – 1 = 0
2x = 1
or
x=
1
2
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
(xvii) 6x2 – 7x – 13 = 0
Sol.
6x2 – 7x – 13 = 0

6x2 – 13x + 6x – 13
 x(6x – 13) + 1(6x – 13)

(6x – 13) (x + 1)

6x – 13 = 0

6x = 13

x=
13
6
(xviii) 3x2 + 34x + 11 = 0
Sol.
3x2 + 34x + 11 = 0

3x2 + 33x + x + 11
 3x (x + 11) + 1 (x + 11)

(x + 11) (3x + 1)

x + 11 = 0

x = – 11

(xix)
Sol.
(xx)
Sol.
(xxi)
Sol.
x = – 11
(3 marks)
=
=
=
or
or
0
0
0
x+1=0
x = –1
or
x = –1
(3 marks)
=
=
=
or
or
0
0
0
3x + 1 = 0
3x = – 1
or
x=
–1
3
(3 marks)
3x2 – 11x + 6 = 0
3x2 – 11x + 6 = 0

3x2 – 9x – 2x + 6

3x(x – 3) – 2(x – 3)

(x – 3) (3x – 2)

x–3=0

x=3
=
=
=
or
or
0
0
0
3x – 2 = 0
3x = 2

or
x=
x=3
2
3
(3 marks)
3x2 – 10x + 8 = 0
3x2 – 10x + 8 = 0

3x2 – 6x – 4x + 8

3x(x – 2) – 4(x – 2)

(x – 2) (3x – 4)

x–2=0

x=2
=
=
=
or
or
0
0
0
3x – 4 = 0
3x = 4

or
x=
x=2
2m2 + 19m + 30 = 0
2m2 + 19m + 30 = 0

2m2 + 15m + 4m + 30
 m(2m + 15) + 2(2m + 15)

(2m + 15) (m + 2)

2m + 15 = 0

2m = –15

S C H O O L S E C TI O N
m=
–15
2
4
3
(3 marks)
=
=
=
or
or
0
0
0
m+2=0
m = –2
or
m = –2
71
MT
ALGEBRA
EDUCARE LTD.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
18.
(i)
Sol.
Solve the following quardratic equations by factorization method.
4y2 + 4y + 1 = 0
(3 marks)
4y2 + 4y + 1 = 0

4y2 + 2y + 2y + 1 = 0
 2y (2y + 1) + 1 (2y + 1) = 0

(2y + 1) (2y + 1) = 0

(2y + 1)2 = 0
Taking square root on both the sides, we get;
2y + 1 = 0

2y = – 1

(iii)
Sol.
(vi)
Sol.
y
=
1
2
(3 marks)
3x2 + 10 = 11x
3x2 + 10 = 11x

3x2 – 11x + 10
2

3x – 6x – 5x + 10

3x (x – 2) – 5 (x – 2)

(x – 2) (3x – 5)

x–2=0

x=2
=
=
=
=
or
or

or x =
x=2
8x2 – 22x – 21 = 0
8x2 – 22x – 21 = 0

8x2 – 28x + 6x – 21
 4x (2x – 7) + 3 (2x – 7)

(2x – 7) (4x + 3)

2x – 7 = 0

2x = 7
7

x=
2
0
0
0
0
3x – 5 = 0
3x = 5
(3 marks)
=
=
=
or
or
0
0
0
4x + 3 = 0
4x = – 3
3
or x =
4
(3 marks)
(vii)
Sol.
9y2 – 3y – 2 = 0
9y2 – 3y – 2 = 0

9y2 – 6y + 3y – 2
 3y (3y – 2) + 1 (3y – 2)

(3y – 2) (3y + 1)

3y – 2 = 0

3y = 2
2

y=
3
(viii)
Sol.
5z2 – 3z – 2 = 0
5z2 – 3z – 2 = 0

5z2 – 5z + 2z – 2

5z (z – 1) + 2 (z – 1)

(z – 1) (5z + 2)

z–1=0

z=1
=
=
=
or
or

or z =
72
z=1
5
3
=
=
=
or
or
0
0
0
3y + 1 = 0
3y = – 1
1
or y =
3
(3 marks)
0
0
0
5z + 2 = 0
5z = – 2
2
5
S C H O O L S E C TI O N
MT
(xi)
Sol.
(xii)
Sol.
=
=
=
=
or
or

or x =
x=2
6y2 + 17y + 12 = 0
6y2 + 1y + 12 = 0

6y2 + 8y + 9y + 12
 2y (3y + 4) + 3 (3y + 4)

(3y + 4) (2y + 3)

3y + 4 = 0

3y = – 4
y=
4
3
0
0
0
0
9x – 4 = 0
9x = 4
4
9
(3 marks)
=
=
=
or
or
0
0
0
2y + 3 = 0
2y = – 3
or y =
3
2
1 2 2
x –
–x+1=0
9
3
Sol.




(3 marks)
1 2 2
x –
–x+1=0
9
3
Multiplying throughout by 9, we get;
x2 – 6x + 9 = 0
2
x – 3x – 3x + 9 = 0
x (x – 3) – 3 (x – 3) = 0
(x – 3) (x – 3) = 0
(x – 3)2 = 0
Taking square root on both the sides, we get;
x–3 = 0

(x)
(3 marks)
9x2 + 8 = 22x
9x2 + 8 = 22x

9x2 – 22x + 8
2

9x – 18x – 4x + 8

9x (x – 2) – 4 (x – 2)

(x – 2) (9x – 4)

x–2=0

x=2

(iv)
ALGEBRA
EDUCARE LTD.
x
5t2 –
Sol.
=
3
17
3
t+
=0
2
2
(3 marks)
17
3
t+
=0
2
2
Multiplying throughout by 2, we get,
10t2 – 17t + 3 = 0
10t2 – 15t – 2t + 3 = 0
5t (2t – 3) – 1 (2t – 3) = 0
(2t – 3) (5t – 1) = 0
2t – 3 = 0 or 5t – 1 = 0
2t = 3 or 5t = 1
5t2 –






S C H O O L S E C TI O N
t=
3
2
or t =
1
5
73
MT
ALGEBRA
(ix)
Sol.
1
1
= 2
x+ 2
x
1
=
x+2




x (x



EDUCARE LTD.
(3 marks)
1
x2
x2
x –x–2
x2 – 2x + x – 2
– 2) + 1 (x – 2)
(x – 2) (x + 1)
x–2=0
2
x=2
=
=
=
=
=
or
x+2
0
0
0
0
x+1=0
or x = – 1
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :
1.
Solve the following quadratic equations by factorization method.
(vi)
x+
Sol.





20
– 12 = 0
x
20
x+
– 12 = 0
x
Multiplying throughout by x, we get,
x2 + 20 – 12x = 0
x2 – 12x + 20 = 0
2
x – 10x – 2x + 20 = 0
x(x – 10) – 2(x – 10) = 0
(x – 10) (x – 2) = 0
x – 10 = 0 or x – 2 = 0

(ix)
x = 10
or x = 2







10
=1
x
10
2x –
=1
x
Multiplying throughout by x, we get,
2x2 – 10 = x
2
2x – x – 10 = 0
2x2 – 5x + 4x – 10 = 0
x(2x – 5) + 2(2x – 5) = 0
(2x – 5) (x + 2) = 0
2x – 5 = 0 or x + 2 = 0
2x = 5 or x = – 2

x=
2x –
Sol.
5
2
(3 marks)
(3 marks)
or x = – 2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
9.
(ii)
Sol.
Solve the following equations
3x – x2 = 0
3x – x2 = 0

0 =
2

x – 3x =

x (x – 3) =

x = 0 or

74
x=0
by factorization method.
(3 marks)
x2 – 3x
0
0
x–3=0
or x = 3
S C H O O L S E C TI O N
MT
(i)
Sol.
ALGEBRA
EDUCARE LTD.
(3 marks)
y2 – 3 = 0
y2 – 3 = 0


(y +

y2 – ( 3 )2
3 ) (y – 3 )
y+ 3 =0
= 0
= 0
or y –
3 =0
y=– 3
or y =
3

(vii)
Sol.
9x2 – 16 = 0
9x2 – 16 = 0

(3x)2 – (4)2

3 (x + 4) (3x – 4)

3x + 4 = 0

3x = – 4
–4
x=
3

(viii)
Sol.
49x2 = 36
49x2 = 36

49x2 – 36

(7x)2 – (6)2

(7x + 6) (7x – 6)

7x + 6 = 0

7x = – 6
–6

x=
7
(3 marks)
= 0
= 0
or 3x – 4 = 0
or 3x = 4
4
or x =
3
(3 marks)
=
=
=
or
or
0
0
0
7x – 6 = 0
7x = 6
6
or x =
7
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :
1.
(v)
Sol.
Solve the following quadratic
m2 – 84x = 0
m2 – 84 = 0
2

(m)2 –
84
=
2
2

(m) –
4  21
=
2
2

(m – 2 21
=
 m  2 21 m – 2 21 =

m  2 21 = 0 or










m = – 2 21
(xxii) 7m2 – 84 = 0
Sol.
7m2 – 84 = 0

7(m2 – 12)

m2 –12


–  4  3
(m)2 –

(m)2

12
2
2

(m + 2 3 ) (m – 2 3 )

m+ 2 3 =0

m = –2 3
S C H O O L S E C TI O N
equations by factorization method.
(3 marks)
0
0
0
0
m – 2 21 = 0
or m = 2 21
(3 marks)
=
=
0
0
7
0
=
0
=
0
=
or m – 2 3 = 0
or m = 2 3
75
MT
ALGEBRA
EDUCARE LTD.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
18.
(v)
Sol.
Solve the following quardratic
(2y + 3)2 = 81
(2y + 3)2 = 81

(2y + 3)2 – 81 =

(2y + 3)2 – (9)2 =
 (2y + 3 + 9) (2y + 3 – 9) =

(2y + 12) (2y – 6) =

2y + 12 = 0 or

2y = – 12 or
–12

y=
or
2

o
y=–6
equations by factorization method.
(3 marks)
0
0
0
0
2y – 6 = 0
2y = 6
6
y=
2
or y = 3
Method 2 : COMPLETING SQUARE METHOD
(Assuming the variable involved in equation to be x)
Step 1 : Check the coefficient of x2, it has to be 1. If not, then make it 1 by
dividing each term by the coefficient of x2.
Step 2 : Keep only the constant term on the R.H.S.
Step 3 : Find the third term, using the formula :
2
Step 4 :
Step 5 :
Step 6 :
Step 7 :
1

Third term =  × coefficient of x 
2
Add third term on both the sides. Due to this, L.H.S. becomes a perfect square.
Express L.H.S. in a square form.
Now take square root on both sides and write the square root of
R.H.S. with '  ' sign.
Find the value of x and write the solution set.
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 46) :
1.
(i)
Sol.
Solve the following quadratic equations by completing square.
x2 + 8x + 9 = 0
(3 marks)
x2 + 8x + 9 = 0
 x2 + 8x = –9
.....(i)
Third term
=
1

  coefficient of x 
2
2
2
1

=   8
2

2
= (4)
= 16
Adding 16 to both sides of (i) we get,
x2 + 8x + 16
= –9 + 16
 (x + 4)2
= 7
 (x + 4)2
=
 7
2
Taking square root on both the sides we get,
x+4
 x
 x=–4+ 7
=
 7
= –4+
7
or x = – 4 –
– 4 + 7 and – 4 –
76
7
7 are the roots of the given quadratic equations.
S C H O O L S E C TI O N
MT
(ii)
Sol.
ALGEBRA
EDUCARE LTD.
(3 marks)
z2 + 6z – 8 = 0
z2 + 6z – 8 = 0
 z2 + 6z = 8
Third term
.... (i)
=
1

  coefficient of z 
2
=
1

  6
2
2
2
= (3) 2
= 9
Adding 9 to both sides of (i), we get,
z2 + 6z + 9
= 8+9
 (z + 3)2
= 17
Taking square root on both the sides we get,
z+3
=
 17
 z
= –3 +
 z = –3 + 17
or z = –3 –
 – 3 + 17 and – 3 –
(iii)
Sol.
m2 – 3m – 1 = 0
m2 – 3m – 1 = 0
 m2 – 3m = 1
Third term
Adding
17
17 are the roots of the given quadratic equations.
(3 marks)
.... (i)
=
1

  coefficient of m
2
=
1

  – 3
2
=
 –3 
 
2
=
9
4
2
2
2
9
to both sides of (i) we get,
4
m2 – 3m +
2

3

 m – 
2
2

3

 m – 
2

17
9
4
9
4
=
1
=
49
4
13
4
Taking square root on both the sides we get,
m–
3
2
 m
S C H O O L S E C TI O N
=
13
2
=

=
3

2
13
2
77
MT
ALGEBRA
 m
 m=

(iv)
Sol.
3  13
2
=
3  13
2
or
m=
EDUCARE LTD.
3 – 13
2
3  13
3 – 13
and
are the roots of the given quadratic equations.
2
2
(3 marks)
y2 = 3 + 4y
y2 = 3 + 4y
 y2 – 4y = 3
.... (i)
Third term
=
1

 × coefficient of y 
2
=
1

 × – 4
2
2
2
=
(– 2)2
=
4
Adding 4 to both sides of (i) we get,
y2 – 4y + 4
=
3+4
 (y – 2)2
=
7
Taking square root on both the sides we get,
y–2
 y
 y=2+
 2 +
(v)
Sol.
7
7 and 2 –
=
 7
=
2+
or
y=2–
7
7
7 are the roots of the given quadratic equations.
(3 marks)
p2 – 12p + 32 = 0
p2 – 12p + 32 = 0
 p2 – 12p = –32
.... (i)
Third term




=
1

  coefficient of p
2
=
1

  – 12
2
2
2
= (– 6)2
= 36
Adding 36 to both sides of (i), we get,
p2 – 12p + 36 = – 32 + 36
(p – 6)2
= 4
Taking square root on both the sides we get,
p–6
= +2
p
= 6+2
p=6+2
or p = 6 – 2
p=8
or p = 4
 8 and 4 are the roots of the given quadratic equations.
78
S C H O O L S E C TI O N
MT
(vi)
Sol.
ALGEBRA
EDUCARE LTD.
(3 marks)
x (x – 1) = 1
x (x – 1) = 1
 x2 – x = 1
Third term
.... (i)
=
1

  coefficient of x 
2
=
1

  – 1
2
=
 –1
 
2
=
1
4
2
2
Adding
1
to both sides of equation (i) we get,
4
x2 – x +
1
4
2

1

 x – 
2
1

 x – 
2
2

= 1+
=
1
4
4 1
4
5
4
Taking square root on both the sides we get,
=
x
=
1
5

2
2
 x
=
1± 5
2
 x=
1
5
+
2
2
or x =
1
5
–
2
2
 x=
1 5
2
or x =
1– 5
2

2
1 5
1– 5
and
are the roots of the given quadratic equations.
2
2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
19.
(i)
Sol.
Solve the following quadratic equation by completing square.
x2 + 3x + 1 = 0
x2 + 3x + 1 = 0
 x2 + 3x = – 1 ..... (i)
Third term
S C H O O L S E C TI O N
=
1

  coefficient of x 
2

=
1

 3
2

2
2
79
MT
ALGEBRA
=
=
Adding
9
to both the sides of (i) we get,
4
x2 + 3x +

3

x 
2

2
3
 
2
9
4
EDUCARE LTD.
2
9
9
=–1+
4
4
49
=
4
2
3
5

 x 
=
2
4


Taking square root on both the sides we get,
3
5
x+
= ±
2
2
3
5
 x
=
+
2
2
 x
 x=

(ii)
Sol.
=
–3 + 5
2
–3 ± 5
2
or x =
–3 + 5
–3–
and
2
2
–3–
2
5
5
are the roots of the given quadratic equation.
(3 marks)
z2 + 4z – 7 = 0
z2 + 4z – 7 = 0
 z2 + 4z = 7
..... (i)
2
=
=
Adding 4 to both
z2 + 4z + 4
=
 (z + 2)2
=
1

 × coefficient of z 
2

2
1

 × 4
2

(2) 2
4
the sides of (i) we get,
7+4
11
 (z + 2)2
( 11)
Third term
=
=
=
2
Taking square root on both the sides, we get;
z+2
=
± 11
 z
=
– 2  11
 z = – 2 + 11
or z = – 2 – 11

80
– 2 + 11 and – 2 – 11 are the roots of the given quadratic equation.
S C H O O L S E C TI O N
MT
(iii)
Sol.
ALGEBRA
EDUCARE LTD.
(3 marks)
n2 + 3n – 4 = 0
n2 + 3n – 4 = 0
 n2 + 3n = 4
......(i)
Third term
=
=
=
=
Adding
1

 × coefficient of n 
2

2
1

 × 3
2

2
3
 
2
2
9
4
9
to both the sides of (i) we get,
4
n2 + 3n +
9
9
=4+
4
4
2
3
16 + 9

 n + 
=
2
4

2
3
25

 n + 
=
2
4

Taking square root on both the sides we get,
n+
3
2
 n
 n=
5
2
=
±
=
–3 5

2
2
3
3
5
5
–
or n =
–
2
2
2
2
2
2
 n=1
 n=
8
2
or n = – 4
or n = –
 1 and – 4 are the roots of the given quadratic equation.
(iv)
Sol.
m2 = 4 + 5m
m2 = 4 + 5m
 m2 – 5m = 4
Third term
(3 marks)
..... (i)
=
1

 × coefficient of m 
2

=
1

 × – 5
2


=
 –5


 2 
=
S C H O O L S E C TI O N
2
2
2
25
4
81
MT
ALGEBRA
Adding
EDUCARE LTD.
25
to both the sides of (i) we get,
4
m2 – 5m +
25
25
= 4+
4
4
2
5
16 + 25

=
m – 
2
4

2
5
41

 m – 
=
2
4

Taking square root on both the sides we get,

m–
41
2
=
±
 m
=
5

2
41
2
 m
=
5
41
 m=

(v)
5
2
5+
41
2
2
or m =
5  41
2
5 + 41
5  41
and
are the roots of the given quadratic equation.
2
2
(3 marks)
p2 – 10p + 5 = 0
Sol.
p2 – 10p + 5 = 0
 p2 – 10p = – 5 ..... (i)
Third term
=
1

 × coefficient of p
2
1

 × –10
2
= (– 5)2
= 25
2
2
=
Adding 25 to both the sides of equation (i), we get;
p2 – 10p + 25
 (p – 5)
82
= – 5 + 25
= 20
 p–5
= +
 p–5
= + 2 5
 p
=
 p= 52 5
or p = 5 – 2 5

(vi)
Sol.
2
20
[Taking square roots on both sides]
52 5
5  2 5 and 5 – 2 5 are the roots of the given quadratic equation.
x (x – 5) = 6
x (x – 5) = 6
 x2 – 5x = 6
(3 marks)
..... (i)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
Third term
Adding
=
1

 × coefficient of x 
2
=
1

 × – 5
2
=
 – 5


2 
=
25
4
2
2
2
25
to both the sides of (i) we get,
4
x 2 – 5x 
25
4
=
6
25
4
2
5
24 + 25

 x – 
=
2
4


2
5
49

 x – 
=
2
4

Taking square root on both the sides, we get;
x
 x=
=
5
7
+
2
2
5 7

2 2
or x =
5
7
–
2
2
2
2
or x = – 1
12
2
 x=6
 x=
or x =
 6 and – 1 are the roots of the given quadratic equation.
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 46) :
1.
(vii)
Sol.
Solve the following quadratic equations by completing square.
3y2 + 7y + 1 = 0
(3 marks)
3y2 + 7y + 1 = 0
 3y2 + 7y = –1
Dividing both the sides by 3 we get,
7
y
3
=
–1
3
Third term
=
1

 × coefficient of y 
2
=
 1 7
 × 
2 3
=
 7
 
6
=
49
36
y2 +
S C H O O L S E C TI O N
.....(i)
2
2
2
83
MT
ALGEBRA
Adding
y2 +
7
49
y+
=
3
36
7

 y + 
6
2
2

7

 y + 
6
=
–1 49
+
3
36
–12 + 49
36
37
36
Taking square root on both the sides, we get,
=
7
6
37
6
=

 y
=
–
 y
=
– 7  37
6
y+
 y=
(viii)
Sol.
49
to both sides of (i) we get,
36


EDUCARE LTD.
7

6
37
6
– 7  37
– 7 – 37
or y =
6
6
– 7  37
– 7 – 37
and
are the roots of the given quadratic equations.
6
6
(3 marks)
4p2 + 7 = 12p
4p2 + 7 = 12p
 4p2 – 12p = –7
Dividing both the sides by 4 we get,
–7
.... (i)
4
We make L.H.S. a perfect square by using
p2 – 3p =
Third term
Adding

84
1

 × coefficient of p
2
=
1

 × – 3
2
=
 – 3


2 
=
9
4
2
2
2
9
to both sides of (i) we get,
4
p2 – 3p +
3

 p – 
2
=
9
4
=
–7
9
+
4
4
=
2
4
2
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
Taking square root on both the sides we get,
p–
3
2
 p
 p
 p=

(ix)
Sol.
2
2
3
2

=
2
2
3 2
=
2
3– 2
or p =
2
=
3 2
2
3 2
3– 2
and
are the roots of the given quadratic equations.
2
2
6m2 + m = 2
6m2 + m = 2
Dividing both the sides by 4 we get,
1
1
.....(i)
m2 + m =
6
3
2
1

Third term
=  × coefficient of m
2

Adding
m2 +

=
 1 1
 × 
2 6
=
 1
 
12
=
1
144
(3 marks)
2
2
1
to both sides of (i) we get,
144
1
1
m+
6
144
1

 m 

12 
=
1
1
+
3
144
=
48  1
144
2
2
1
49

 m 
=


12 
144
Taking square root on both the sides we get,
1
12
=

 m=
–1
7
+
12
12
or
m=
–1
7
–
12
12
 m=
6
12
or
m=
8
12
 m=
1
2
or
m=
2
3
m

7
12
1
2
and
are the roots of the given quadratic equations.
2
3
S C H O O L S E C TI O N
85
MT
ALGEBRA
EDUCARE LTD.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
19.
(vii)
Sol.
Solve the following quadratic equations by completing square :
2y2 + 5y + 1 = 0
(3 marks)
2y2 + 5y + 1 = 0
 2y2 + 5y = – 1
Dividing both the sides by 2 we get,
5
y
2
=
1
.....(i)
2
Third term
=
1

 × coefficient of y 
2

=
1 5
 × 
2 2
=
5
 
4
=
25
16
y2 +
Adding
y2 +
2
2
2
25
to both the sides of (i) we get,
16
5
25
y+
=
2
16
1
25
+
2
16
2
5
– 8 + 25

=
y + 
4
16

2
5
17

 y + 
=
4
16

Taking square root on both the sides, we get;

17
2
±
 y
=
–5

4
` y
=
–5  17
4
 y=

(viii)
Sol.
5
4
=
y+
– 5 + 17
4
or y =
17
4
– 5 – 17
4
– 5 + 17
– 5 – 17
and
are the roots of the given quadratic equation.
4
4
3p2 + 4 = – 7p
3p2 + 4 = – 7p
 3p2 + 7p = – 4
Dividing both the sides by 3 we get,
(3 marks)
7
4
p=
.....(i)
3
3
We make L.H.S. a perfect square by using
p2 +
86
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
Third term
=
=
=
=
Adding
2
49
36
49
to both the sides of (i) we get,
36
7
49
p+
=
3
36
p2 +
1

 × coefficient of p 
2

2
1 7
 × 
2 3
2
7
 
6
4
49
+
3
36
2
7
– 48 + 49

 p + 
=
6
36

Taking square root on both the sides, we get;
7
6
±
 p
=
–7 1

6
6
7
1
+
6
6
6
 p=–
6
7
1
–
6
6
8
or p = –
6
4
or p = –
3
 p=–
 p=–1
 – 1 and –
(ix)
Sol.
1
6
=
p+
or p = –
4
are the roots of the given quadratic equation.
3
5m2 + m = 3
5m2 + m = 3
Dividing both the sides by 5 we get,
1
3
.... (i)
m2 + m =
5
5
Third term
Adding
S C H O O L S E C TI O N
=
1

 × coefficient of m 
2

=
1 1
 × 
2 5
=
 1 


 10 
=
1
100
2
2
2
1
to both the sides of (i) we get,
100
87
MT
ALGEBRA
m2 +
1
1
m+
5
100
EDUCARE LTD.
3
1
+
5
100
=
2
1 
60 + 1

=
m +

10 
100

2
1 
61

 m +
=

10 
100

Taking square root on both the sides we get,

 m
=
–1
61

10
10
 m
=
–1  61
10
or
m=

Sol.
61
10
±
 m=
o
1
10
=
m+
–1 + 61
10
–1 – 61
10
–1 + 61
–1 – 61
and
are the roots of the given quadratic equation.
10
10
Method 3 : FORMULA METHOD
An ancient Indian mathematician Shridharacharya arond 1025 A.D. derived
a formula for solving quadratic equations. So this formula is known as
Shridharacharya fromula for finding the roots of the quadratic equations
ax2 + bx + c = 0
Derivation of Formula for Formula Method :
ax2 + bx + c = 0
ax2 + bx + c = 0
Dividing throught by 'a', we get
ax 2
bx
c


= 0
a
a
a
c
b
c
2
 x  x–
= 
a
a
a
2
1

Third Term
=  × coefficient of x 
2

=
 1 b
 × 
2 a
=
 b


2a 
=
b2
4a 2
2
2
b2
on both the sides we get,
4a 2
b
b2
c
b2
x2  x 
–

=
a
4a 2
a 4a 2
2
b
– 4ac  b2


x
=


2a 
4a 2
Adding


88
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.

b

 x 

2a 

x
2
b
2a
 x
 x
=
b2 – 4ac
4a 2
=

=
–
=
b2 – 4ac
[Taking square roots on both the sides]
4a 2
b

2a
–b 
b2 – 4ac
2a
b2 – 4ac
2a
–b 
b2 – 4ac
is refered as quadratic formula for solving quadratic
2a
equation. The roots of the quadratic equation are denoted as  and .
 x=
  =
o
–b 
b2 – 4ac
2a
, =
–b –
b2 – 4ac
2a
 – b  b2 – 4ac – b – b2 – 4ac 
,

 Solution set = 
2a
2a


Steps involved in Formula Method :
(Assuming the variable involved in equation to be x)
Step 1 : Write the equation in the standard form.
Step 2 : Comparing the given equation with standard form
i.e. ax² + bx + c = 0, write the value of a, b and c.
Step 3 : Find the value of b2 - 4ac.
Step 4 : Find the value of x using formula : x =
–b 
b2 – 4ac
2a
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
20.
(i)
Sol.
Solve the following quadratic equations using formula.
x2 + 3x – 10 = 0
(3 marks)
x2 + 3x – 10 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = – 10
= (3)2 – 4 (1) (– 10)
b2 – 4ac
= 9 + 40
= 49
x
=
=
–b ±
b2 - 4ac
2a
– 3 ± 49
2 (1)
–3 ± 7
2
–3 + 7
–3 – 7
 x=
or x =
2
2
4
–10
 x=
or x =
2
2
 x = 2 or x = – 5
=
 2 and – 5 are the roots of given quadratic equation.
S C H O O L S E C TI O N
89
MT
ALGEBRA
EDUCARE LTD.
EXERCISE - 2.5 (TEXT BOOK PAGE NO. 49) :
1.
(i)
Sol.
Solve the following quadratic equations by using formula.
m2 – 3m – 10 = 0
(3 marks)
2
m – 3m – 10 = 0
Comparing with am2 + bm + c = 0 we have a = 1, b = –3, c = –10
b2 – 4ac
= (– 3)2 – 4 (1) (–10)
= 9 + 40
= 49
m
=
=
–b ±
b2 – 4ac
2a
– (– 3) ± 49
2 (1)
37
2
3–7
or m =
2
–4
or m =
2
or m = –2
=
37
2
10
 m =
2
 m = 5
 m =
 5 and –2 are the roots of the given quadratic equation.
(ii)
Sol.
x2 + 3x – 2 = 0
(3 marks)
x2 + 3x – 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = –2
b2 – 4ac
= (3)2 – 4(1) (–2)
= 9+8
= 17
x
 x=

=
– 3 + 17
2
–b ±
b2 – 4ac
2a
=
– 3 ± 17
2 (1)
=
– 3 ± 17
2
or x =
– 3 – 17
2
– 3 + 17
– 3 – 17
and
are the roots of the given quadratic equation.
2
2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
20.
(ii)
Sol.
90
Solve the following quadratic equations using formula.
2x2 + 5x – 2 = 0
(3 marks)
2x2 + 5x – 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 2, b = 5, c = – 2
b2 – 4ac
= (5)2 – 4 (2) (– 2)
= 25 + 16
= 41
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
x
 x=

=
– 5 + 41
4
–b ±
b2 – 4ac
2a
=
– 5 ± 41
2(2)
=
– 5 ± 41
4
or x =
– 5 – 41
4
– 5 + 41
– 5 – 41
and
are the roots of the given quadratic equation.
4
4
EXERCISE - 2.5 (TEXT BOOK PAGE NO. 49) :
1.
(iv)
Sol.
Solve the following quadratic equations by using formula :
5m2 – 2m = 2
(3 marks)
5m2 – 2m = 2
 5m2 – 2m – 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 5, b = – 2, c = – 2
b2 – 4ac
= (– 2)2 – 4 (5) (– 2)
= 4 + 40
= 44
m
=
=
=
=
=
=
 x=

(v)
Sol.
1  11
5
–b 
b2 – 4ac
2a
– (– 2)  44
2 (5)
2
4  11
10
2  2 11
10

2 1  11

10
1  11
5
or x =
1 – 11
5
1  11
1 – 11
and
are the roots of the given quadratic equation.
5
5
7x + 1 = 6x2
(3 marks)
 0 = 6x2 – 7x – 1
Comparing with ax2 + bx + c = 0 we have a = 6, b = – 7, c = – 1
b2 – 4ac
= (– 7)2 – 4 (6) (– 1)
= 49 + 24
= 73
S C H O O L S E C TI O N
91
MT
ALGEBRA
x
EDUCARE LTD.
–b 
b2 – 4ac
2a
– (– 7)  73
2 (6)
=
=
7  73
12
7 – 73
or x =
12
=
 x=

(vi)
Sol.
7  73
12
7  73
7 – 73
and
are the roots of the given quadratic equation.
12
12
2x2 – x – 4 = 0
(3 marks)
 2x2 – x – 4 = 0
Comparing with ax2 + bx + c = 0 we have a = 2, b = – 1, c = – 4
b2 – ac
= (– 1)2 – 4 (2) (– 4)
= 1 + 32
= 33
x
–b 
=
– (–1)  33
2 (2)
=

(vii)
Sol.
1
33
4
1
33
4
and
1
33
4
1 – 33
or x =
4
=
 x=
b2 – 4ac
2a
1–
33
4
are the roots of the given quadratic equation.
3y2 + 7y + 4 = 0
 3y2 + 7y + 4 = 0
Comparing with ax2 + bx + c = 0 we have a = 3, b = 7, c = 4
b2 – ac
= (7)2 – 4 (3) (4)
= 49 – 48
= 1
x
=
=
(3 marks)
–b 
b2 – 4ac
2a
–7 1
2 (3)
– 7 1
6
–7 – 1
or y =
6
–4
or y =
3
=
 y=
–7  1
6
 y=–1
 – 1 and
92
–4
are the roots of the given quadratic equation.
3
S C H O O L S E C TI O N
MT
(viii)
Sol.
ALGEBRA
EDUCARE LTD.
2n2 + 5n + 2 = 0
 2n2 + 5n + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 2, b = 5, c = 2
b2 – 4ac
= (5)2 – 4 (2) (2)
= 25 – 16
= 9
n
b2 – 4ac
2a
=
–5  9
2 (2)
=
–5  3
4
–5  3
4
or n =
–5 – 3
4
 n=
–2
4
or n =
–8
4
 n=
–1
2
or n = – 2
–1
and – 2 are the roots of the given quadratic equation.
2
7p2 – 5p – 2 = 0
(3 marks)
2
 7p – 5p – 2 = 0
Comparing with ap2 + bp + c = 0 we have a = 7, b = – 5, c = – 2
b2 – 4ac
= (– 5)2 – 4 (7) (– 2)
= 25 + 56
= 81
p
=
 p=
59
14
 p=
14
14
 p=1
 1 and
(x)
Sol.
–b 
 n=

(ix)
Sol.
=
(3 marks)
b2 – 4ac
2a
=
– (– 5)  81
2 (7)
=
59
14
or p =
5–9
14
–4
14
–2
or p =
7
or p =
–2
are the roots of the given quadratic equation.
7
9s2 – 4 = – 6s
9s2 – 4 = – 6s
 9s2 + 6s – 4 = 0
S C H O O L S E C TI O N
–b 
(3 marks)
93
MT
ALGEBRA
EDUCARE LTD.
Comparing with as2 + bs + c = 0 we have a = 9, b = 6, c = – 4
b2 – ac
= (6)2 – 4 (9) (– 4)
= 36 + 144
= 180
– b  b2 – 4ac
s
=
2a
– 6  180
=
2 (9)
=
=
=
=
 s=

(xi)
Sol.
–1  5
3
–6 
36  5
18
–6  6 5
18

6 –1 
5

18
–1  5
3
or s =
–1 – 5
3
–1  5
–1 – 5
and
are the roots of the given quadratic equation.
3
3
3q2 = 2q + 8
(3 marks)
3q2 = 2q + 8
 3q2 – 2q – 8 = 0
Comparing with aq2 + bq + c = 0 we have a = 3, b = – 2, c = – 8
b2 – 4ac
= (– 2)2 – 4 (3) (– 8)
= 4 + 96
= 100
q
=
=
–b 
b2 – 4ac
2a
– (– 2)  100
2 (3)
2  10
6
2 – 10
or q =
6
–8
or q =
6
–4
or q =
3
=
2  10
6
12
 q=
6
 q=
 q=2
 2 and
(xii)
Sol.
94
–4
are the roots of the given quadratic equation.
3
4x2 + 7x + 2 = 0
4x2 + 7x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = 7, c = 2
(3 marks)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
b2 – 4ac
= (7)2 – 4 (4) (2)
= 49 – 32
= 17
x
=
 x=

– 7  17
8
–b 
b2 – 4ac
2a
=
– 7  17
2 (4)
=
– 7  17
8
– 7 – 17
8
or x =
– 7  17
– 7 – 17
and
are the roots of the given quadratic equation.
8
8
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
20.
(iv)
Sol.
Solve the following quadratic equations using formula.
6m2 – 4m = 3
(3 marks)
6m2 – 4m = 3
 6m2 – 4m – 3 = 0
Comparing with am2 + bm + c = 0, we have a = 6, b = – 4, c = – 3
b2 – 4ac
= (– 4)2 – 4 (6) (– 3)
= 16 + 72
= 88
m
=
=
=
=
=
=
 m=

(v)
Sol.
2 + 22
6
b2 - 4ac
2a
– (– 4) ± 88
–b ±
2(6)
4 ± 4 × 22
12
4 ± 2 22
12

2 2
22

12
2±
22
6
or m =
2–
22
6
2 + 22
2 – 22
and
are the roots of the given quadratic equation.
6
6
9x + 1 = 4x2
(3 marks)
9x + 1 = 4x2
 0 = 4x2 – 9x – 1
 4x2 – 9x – 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = – 9, c = – 1
S C H O O L S E C TI O N
95
MT
ALGEBRA
b2 – 4ac
= (– 9)2 – 4 (4) (– 1)
= 81 + 16
= 97
x
=
=
=
9 + 97
 x=
8

(vi)
Sol.
2(4)
9±
or x =
97
8
9–
97
8
4x2 + x – 5 = 0
(3 marks)
4x2 + x – 5 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = 1, c = – 5
b2 – 4ac
= (1)2 – 4 (4) (– 5)
= 1 + 80
= 81
=
–b ±
b2 – 4ac
2a
=
–1 ± 81
2(4)
=
–1 ± 9
8
 x=
–1 + 9
8
or x =
–1 – 9
8
 x=
8
8
or x =
–10
8
 x=1
or x =
5
4
 1 and
5
are the roots of the given quadratic equation.
4
3y2 + 8y + 5 = 0
3y2 + 8y + 5 = 0
Comparing with ay2 + by + c = 0 we have a = 3, b = 8, c = 5
= (8)2 – 4 (3) (5)
b2 – 4ac
= 64 – 60
= 4
y
96
b2 – 4ac
2a
– (– 9) ± 97
–b ±
9 + 97
9 – 97
and
are the roots of the given quadratic equation.
8
8
x
(vii)
Sol.
EDUCARE LTD.
(3 marks)
b2 – 4ac
2a
–8 ± 4
=
2 (3)
–8 ± 2
=
6
=
–b ±
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
–8 + 2
6
6
 y=
6
 y=
 y=–1
 – 1 and
–8 – 2
6
– 10
or y =
6
- 5
or y =
3
or y =
5
are the roots of the given quadratic equation.
3
(viii)
Sol.
2n2 – 11n + 9 = 0
(3 marks)
2n2 – 11n + 9 = 0
Comparing with an2 + bn + c = 0 we have a = 2, b = – 11, c = 9
b2 – 4ac
= (– 11)2 – 4 (2) (9)
= 121 – 72
= 49
– b ± b2 – 4ac
n
=
2a
– (–11) ± 49
=
2 (2)
11± 7
=
4
11+7
11  7
 n=
or n =
4
4
18
4
 n=
or n =
4
4
9
 n=
or n = 1
2
9

and 1 are the roots of the given quadratic equation.
2
(ix)
Sol.
9p2 – 5p – 4 = 0
(3 marks)
9p2 – 5p – 4 = 0
Comparing with ap2 + bp + c = 0 we have a = 9, b = – 5, c = – 4
b2 – 4ac
= (– 5)2 – 4 (9) (– 4)
= 25 + 144
= 169
p
=
=
=
5 + 13
18
18
 p=
18
 p=
 p=1
 1 and
S C H O O L S E C TI O N
or
or
or
b2 – 4ac
2a
– (– 5) ± 169
–b ±
2 (9)
5 ± 13
18
5 – 13
p=
18
–8
p=
18
4
p=
9
4
are the roots of the given quadratic equation.
9
97
MT
ALGEBRA
EDUCARE LTD.
EXERCISE - 2.5 (TEXT BOOK PAGE NO. 49) :
1.
(iii)
Sol.
Solve the following quadratic equations by using formula :
x –1
x2 +
=0
(3 marks)
3
x –1
x2 +
=0
3
Multiplying throughout by 3 we get,
3x2 + x – 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 3, b = 1, c = – 1
b2 – 4ac
= (1)2 – 4 (3) (– 1)
= 1 + 12
= 13
– b  b2 – 4ac
x
=
2a
–1  13
=
2 (3)
 x=

–1  13
6
=
–1  13
6
or
–1 – 13
6
–1  13
–1 – 13
and
are the roots of the given quadratic equation.
6
6
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
20.
Solve the following quadratic equations using formula.
(iii)
2x2 +
x –1
=0
5
x –1
=0
5
Multiplying throughout by 5 we get,
10x2 + x – 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 10, b = 1, c = – 1
b2 – 4ac
= (1)2 – 4 (10) (– 1)
= 1 + 40
= 41
– b ± b2 – 4ac
x
=
2a
Sol.
2x2 +
 x=

98
(3 marks)
–1 + 41
20
=
–1 ± 41
2(10)
=
–1 ± 41
20
or x =
–1 – 41
20
–1 + 41
–1 – 41
and
are the roots of the given quadratic equation.
20
20
S C H O O L S E C TI O N
MT
o
EDUCARE LTD.
ALGEBRA
Nature of roots of a quadratic equation :
We know that the roots of the quadratic equation ax2 + bx + c = 0, a  0 are
b2 – 4ac
– b – b2 – 4ac
, 
2a
2a
Here the nature of the roots of the quadratic equation is determined by the
value of b2 – 4ac, which is called as discriminant of the quadratic equation
and it is denoted by  (Delta).

(1)
–b 
If b2 – 4ac > 0 then b2 – 4ac is positive real and therefore roots of equation
are real and unequal.
b2 – 4ac
– b – b2 – 4ac
, 
2a
2a
If b2 – 4ac = 0 then b2 – 4ac = 0 and therefore roots of equation are real
–b
–b
and equal so  =
and  =
2a
2a
both the roots are real and equal, we say that the equation has repeated roots.
e.g.  
(2)
i.e.
(3)
–b 
If b2 – 4ac < 0 then b2 – 4ac is not real number and therefore equation
does not have real roots.
Note : If b2 – 4ac > 0 and b2 – 4ac is not a perfect square then the
roots of the quadratic equation are irrational and occur in pair. They
are conjugate of each other.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
5.
(i)
Sol.
Find the values of discriminant of each of the following equation.
x2 – 3x + 2 = 0
(1 mark)
2
x – 3x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = – 3, c = 2
 = b2 – 4ac
= (– 3)2 – 4 (1) (2)
= 9–8
= 1
 
(ii)
Sol.
2x2 + x + 1 = 0
2x2 + x + 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 2, b = 1, c = 1
 = b2 – 4ac
= (– 1)2 – 4 (2) (1)
= 1–8
= –7
 
(iii)
Sol.
= 1
(1 mark)
= –7
x2 – 6x + 7 = 0
(1 mark)
x2 – 6x + 7 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = – 6, c = 7
 = b2 – 4ac
= (– 6)2 – 4 (1) (7)
= 36 – 28
= 8
 
= 8
S C H O O L S E C TI O N
99
MT
ALGEBRA
EDUCARE LTD.
EXERCISE - 2.6 (TEXT BOOK PAGE NO. 52) :
1.
(i)
Sol.
Find the value of discriminant of each of the following equations :
x2 + 4x + 1 = 0
(1 mark)
x2 + 4x + 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 4, c = 1
 = b2 – 4ac
= (4)2 – 4 (1) (1)
= 16 – 4
= 12
 
(ii)
Sol.
3x2 + 2x – 1 = 0
(1 mark)
3x2 + 2x – 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 3, b = 2, c = – 1
 = b2 – 4ac
= (2)2 – 4 (3) (– 1)
= 4 + 12
= 16
 
(iii)
Sol.
Sol.
= 16
x2 + x + 1 = 0
x2 + x + 1 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 1, c = 1
 = b2 – 4ac
= (1)2 – 4 (1) (1)
= 1–4
= –3
 
(iv)
= 12
= –3
3 x2 + 2 2 x – 2 3 = 0
3x  2 2x – 2 3  0
Comparing with ax2 + bx + c = 0 we have a =
 = b2 – 4ac

 

2
 
100
3 , b = 2 2 , c = –2 3

= 32
4x2 + kx + 2 = 0
4x2 + kx + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = k, c = 2
 = b2 – 4ac
= (k)2 – 4 (4) (2)
= k2 – 32
 
(vi)
Sol.
(1 mark)
2
= 2 2 – 4 3 –2 3
= (4 × 2) + (8 × 3)
= 8 + 24
= 32
(v)
Sol.
(1 mark)
(1 mark)
= k2 – 32
x2 + 4x + k = 0
x2 + 4x + k = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 4, c = k
(1 mark)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.

 
= b2 – 4ac
= (4)2 – 4 (1) (k)
= 16 – 4k
= 16 – 4k
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :
8.
(i)
Find the value of discrimininant of each of the following equation.
(1 mark)
3 x2 + 2 2 x – 2 3 = 0
3 x2 + 2 2 x – 2 3 = 0
Sol.
Comparing with ax2 + bx + c = 0 we have a =
 = b2 – 4ac


2
3 , b = 2 2 , c = 2 3
   2 3 
= 2 2 4 3
= 4×2+8×3
= 8 + 24
= 32
 
(ii)
Sol.
4x2 – kx + 2 = 0
(1 mark)
4x2 – kx + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = – k, c = 2
 = b2 – 4ac
= (– k)2 – 4 (4) (2)
= k2 – 32
 
(iii)
Sol.
= 32
= k2 – 32
x2 + 4x + k = 0
x2 + 4x + k = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 4, c = k
 = b2 – 4ac
= (4)2 – 4 (1)k
= 16 – 4k
 
(1 mark)
= 16 – 4k
EXERCISE - 2.6 (TEXT BOOK PAGE NO. 52) :
2.
(i)
Sol.
(ii)
Sol.
Determine the nature of the roots of the following equations from their
discriminants :
(1 mark)
y2 – 4y – 1 = 0
y2 – 4y – 1 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = – 4, c = – 1
 = b2 – 4ac
= (– 4)2 – 4 (1) (-1)
= 16 + 4
= 20
 > 0
Hence roots of the quadratic equation are real and unequal.
y2 + 6y – 2 = 0
(1 mark)
2
y + 6y – 2 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = 6, c = – 2
 = b2 – 4ac
= (6)2 – 4 (1) (– 2)
S C H O O L S E C TI O N
101
MT
ALGEBRA
=
=
 >
Hence
EDUCARE LTD.
36 + 8
44
0
roots of the quadratic equation are real and unequal.
(iii)
Sol.
y2 + 8y + 4 = 0
(1 mark)
y2 + 8y + 4 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = 8, c = 4
 = b2 – 4ac
= (8)2 – 4 (1) (4)
= 64 -16
= 48
 > 0
Hence roots of the quadratic equation are real and unequal.
(iv)
Sol.
2y2 + 5y – 3 = 0
(1 mark)
2y2 + 5y – 3 = 0
Comparing with ay2 + by + c = 0 we have a = 2, b = 5, c = – 3
 = b2 – 4ac
= (5)2 – 4 (2) (– 3)
= 25 + 24
= 49
 > 0
Hence roots of the quadratic equation are real and unequal.
(v)
Sol.
3y2 + 9y + 4 = 0
(1 mark)
3y2 + 9y + 4 = 0
Comparing with ay2 + by + c = 0 we have a = 3, b = 9, c = 4
 = b2 – 4ac
= (9)2 – 4 (3) (4)
= 81 – 48
= 33
 > 0
Hence roots of the quadratic equation are real and unequal.
(vi)
2x2 + 5 3 x + 16 = 0
Sol.
(1 mark)
2x2 + 5 3 x + 16 = 0
Comparing with ax2 + bx + c = 0 we have a = 2, b = 5 3 , c = 16
 = b2 – 4ac
=
=
=
=
 <
Hence
(5 3 )2 – 4 (2) (16)
25 × 3 – 128
75 – 128
– 53
0
roots of the quadratic equation are not real.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
12.
(i)
Sol.
102
Determine the nature of roots of the following equations from their
discriminants.
y2 – 5y + 11 = 0
(1 mark)
y2 – 5y + 11 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = – 5, c = 11
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.

=
=
=
=
  <
Hence
b2 – 4ac
(– 5)2 – 4 (1) (11)
25 – 44
– 19
0
roots of the quadratic equation are not real.
(1 mark)
(ii)
Sol.
y2 + 6y + 9 = 0
y2 + 6y + 9 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = 6, c = 9
 = b2 – 4ac
= (6)2 – 4 (1) (9)
= 36 – 36
= 0
  = 0
Hence roots of the quadratic equation are real and equal.
(iii)
Sol.
y2 + 8y + 5 = 0
(1 mark)
y2 + 8y + 5 = 0
Comparing with ay2 + by + c = 0 we have a = 1, b = 8, c = 5
 = b2 – 4ac
= (8)2 – 4 (1) (5)
= 64 – 20
= 44
  > 0
Hence roots of the quadratic equation are real and unequal.
(iv)
Sol.
2y2 – 7y – 3 = 0
(1 mark)
2y2 – 7y – 3 = 0
Comparing with ay2 + by + c = 0 we have a = 2, b = – 7, c = – 3
 = b2 – 4ac
= (– 7)2 – 4 (2) (– 3)
= 49 + 24
= 73
  > 0
Hence roots of the quadratic equation are real and unequal.
(v)
Sol.
2y2 + 11y – 7 = 0
(1 mark)
2
2y + 11y – 7 = 0
Comparing with ay2 + by + c = 0 we have a = 2, b = 11, c = – 7
 = b2 – 4ac
= (11)2 – 4 (2) (– 7)
= 121 + 56
= 177
  > 0
Hence roots of the quadratic equation are real and unequal.
(vi)
x2 + 3 2 x – 8 = 0
(1 mark)
2
x + 3 2x – 8 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 3 2 , c = – 8
 = b2 – 4ac
Sol.
=
=
S C H O O L S E C TI O N
(3 2 )2 – 4 (1) (– 8)
9 × 2 + 32
103
MT
ALGEBRA
=
=
  >
Hence
EDUCARE LTD.
18 + 32
50
0
roots of the quadratic equation are real and unequal.
EXERCISE - 2.6 (TEXT BOOK PAGE NO. 52) :
3.
(i)
Sol.
Find the value of k for which given equation has real and equal roots :
(k – 12)x2 + 2 (k – 12)x + 2 = 0
(3 marks)
2
(k – 12)x + 2 (k – 12)x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = k – 12, b = 2 (k – 12), c = 2
 = b2 – 4ac
= [2 (k – 12)]2 – 4 (k -12) (2)
= (2k – 24)2 – 8 (k – 12)
= 4k2 – 96k + 576 – 8k + 96
= 4k2 – 104k + 672
 The roots of given equation are real and equal.
  must be zero.

4k2 – 104k + 672 = 0

4 (k2 – 26k + 168) = 0

k2 – 14k – 12k + 168 = 0

k (k – 14) – 12 (k – 14) = 0
 (k – 14) (k – 12) = 0
 k – 14 = 0
or
k – 12 = 0
 k = 14
(ii)
Sol.
or
k = 12
k2x2 – 2 (k – 1)x + 4 = 0
(3 marks)
2 2
k x – 2 (k – 1)x + 4 = 0
Comparing with ax2 + bx + c = 0 we have a = k2, b = – 2 (k – 1), c = 4
 = b2 – 4ac
= [– 2 (k – 1)]2 – 4 (k2) (4)
= (– 2k + 2)2 – 16k2
= 4k2 – 8k + 4 – 16k2
= – 12k2 – 8k + 4
 The roots of given equation are real and equal.
  must be zero.

– 12k2 – 8k + 4
= 0

– 4 (3k2 + 2k – 1)
= 0
0

3k2 + 3k – k – 1
=
- 4

3k (k + 1) – 1 (k + 1) = 0

(k + 1) (3k – 1)
= 0

k+1=0
or 3k – 1 = 0

k=–1
or 3k = 1

k=–1
or k =
1
3
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
13.
Sol.
104
Find m, if the roots of the quadratic equation (m – 1) x2 – 2 (m – 1) x + 1 = 0
has real and equal roots.
(3 marks)
(m – 1) x2 – 2 (m – 1) x + 1 = 0
Comparing with ax2 + bx + c = 0 we have a = m – 1, b = – 2 (m – 1), c = 1
 = b2 – 4ac
= [– 2 (m – 1)]2 – 4 (m – 1) (1)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.









= (– 2m + 2)2 – 4m + 4
= 4m2 – 8m – 4m + 4
= 4m2 – 12m + 8
The roots of the given equation are real and equal.
 must be zero.
4m2 – 12m + 8
= 0
4 (m2 – 3m + 2)
= 0
0
m2 – 3m + 2
=
4
m2 – 2m – m + 2
= 0
m (m – 2) – 1 (m – 2)
= 0
(m – 2) (m – 1)
= 0
m–2=0
or m – 1 = 0

14.
Sol.
m=2
Find c, if the roots of the quadratic equation x2 – 2 (c + 1) x + c2 = 0 has
real and equal roots.
(3 marks)
x2 – 2 (c + 1) x + c2 = 0
Comparing with Ax2 + Bx + C = 0 we have A = 1, B = – 2 (c + 1), C = c2
 = B2 – 4AC
= [– 2 (c + 1)]2 – 4 (1) (c2)
= (– 2)2 (c + 1)2 – 4c2
= 4 (c2 + 2c + 1) – 4c2
= 4c2 + 8c + 4 – 4c2
= 8c + 4
 The roots of the given equation are real and equal.
  must be zero.

8c + 4
=
0

8c
=
–4
–4

c
=
8

o
or m = 1
c
=
–1
2
Relation between the roots and coefficients :
If a and b are the roots of the quadratic equation ax2 + bx + c = 0
–b
c
and ab =
then  +  =
a
a
Let us verify the above relations.
We have the roots of the quadratic equation ax2 + bx + c = 0

–b 
 +  =
=
=
b2 – 4ac
– b – b2 – 4ac
and  
2a
2a
–b 
b2 – 4ac
– b – b2 – 4ac

2a
2a
– 2b
2a
–b
a
Sum of the roots =
S C H O O L S E C TI O N
– coefficient of x
–b
=
coefficient of x 2
a
105
MT
ALGEBRA
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Also,
 =
=
=
=
=
=
–b 
b2 – 4ac – b – b2 – 4ac

2a
2a
 – b
–
2

b2 – 4ac
4a 2
b2 – (b2 – 4ac)
4a 2
b2 – b2  4ac
4a 2
4ac
4a 2
c
a
 Product of the roots =

2
co ns tan t term
c
=
coefficient of x 2
a
EXERCISE - 2.7 (TEXT BOOK PAGE NO. 54) :
1.
Sol.
If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the
other, find k.
(3 marks)
kx2 – 5x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = k, b = – 5, c = 2
Let  and  be the roots of given quadratic equation.
 = 4
[Given]
- b
(5)
5
 +=
=
=
a
k
k
5

4 + 
=
k
5

5
=
k
5
1


=
×
k
5
1


=
k
Also  =
c
a
=

4.
=

4 2
=
2




106
 1
4  
 k
1
4× 2
k
1
4×
2
k
=
=
=
2
k
2
k
2
k
2
k
2
k
k2
k
[  = 4]
1

   k 


= 2
S C H O O L S E C TI O N
MT
2.
Sol.
ALGEBRA
EDUCARE LTD.
Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are in the
ratio 2 : 5.
(3 marks)
x2 + kx + 40 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = k, c = 40
Let a and b be the roots of given quadratic equation.
Ratio of  to  is 2 : 5
[Given]
Considering common multiple as m.
 = 2m and  = 5m
–b
a
2m + 5m
7m
k
–k
=–k
1
= –k
= –k
= – 7m
 +=







 k = – 14
3.
Sol.
c

[  = 2m and  = 5m]
......(i)
40
= 40
1
2m × 5m
= 40
[  = 2m and  = 5m]
10m2
= 40
40
m2
=
10
m2
= 4
Taking square root on both the sides we get,
m
= +2
k
= – 7m
[From (i)]
k = – 7 (2)
or k = – 7 (– 2)
[ m = + 2]
Also . =


=
=
or k = 14
Find k, if one of the roots of the quadratic equation kx2 – 7x + 12 = 0 is 3.
(3 marks)
kx2 – 7x + 12 = 0
Comparing with ax2 + bx + c = 0 we have a = k, b = – 7, c = 12
Let  and  be the roots of given quadratic equation.
 = 3
[Given]
Alternative method :
–b
–  – 7
7
 3 is one of the root of the
 +=
=
=
a
quadratic equation,
k
k
7
kx2 – 7x + 12 = 0

3+
=
[ a = 3]
 x = 3 satisfies the equation,
k
Substituting x = 3 in equation
7


=
– 3 ......(i)
we get,
k
k (3)2 – 7 (3) + 12 = 0
c
12
Also .
=
=

9k – 21 + 12 = 0
a
k

9k – 9
= 0
12

3×
=
[ a = 3] 
9k
= 9
k
9
12
1

k
=
9


=
×
k
3

k
= 1
4


=
..... (ii)
k

7
4
–3
=
[From (i) and (ii)]
k
k
Multiplying throughout by k we get,
S C H O O L S E C TI O N
107
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7 – 3k = 4
– 3k = 4 – 7
– 3k = – 3
–3

k = –3



k
= 1
4.
If the roots of the equation x2 + px + q = 0 differ by 1, prove that p2 = 1 + 4q.
(3 marks)
2
Proof :
x + px + q = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = p, c = q
Let  and  be the roots of given quadratic equation.
–=1
......(i)
[Given]
b
p
  +  =
=
=–p
......(ii)
a
1
c
q
=
......(iii)
Also  =
a
1
We know that,
( – )2 = ( + )2 – 4
 (1) 2
= (– p)2 – 4 (q)
 1
= p2 – 4q
 1 + 4q = p2
 p2
= 1 + 4q
Hence proved.
5.
Sol.
Find k, if the sum of the roots of the quadratic equation 4x2 + 8kx + k + 9 = 0
is equal to their product.
(3 marks)
4x2 + 8kx + k + 9 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = 8k, c = k + 9
Let  and  be the roots of given quadratic equation.
 +  = .
......(i)
[Given]
–b
– 8k
 +=
=
= – 2k
......(ii)
a
4
c
k+9
=
......(iii)
Also  =
a
4

 +  = .
[From (i)]
k+9

– 2k =
[From (ii) and (iii)]
4
 – 2k × 4 = k + 9

– 8k = k + 9
 – 8k – k = 9

– 9k = 9
9

k =
9

k = –1

6.
Sol.
108
k = –1
If  and  are the roots of the equation x2 – 5x + 6 = 0, find
 
(i) 2 + 2 (ii)  + 
(4 marks)
2
x – 5x + 6 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = -5, c = 6
 and  be the roots of given quadratic equation. [Given]
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
 +=
–

=
– (– 5)
1
=5
c
6
=
=6
a
1
2
2
 + 
= ?
(i)
We know that,
2 + 2 = ( + )2 – 2,
= (5)2 – 2 (6)
= 25 – 12
= 13
Also .

(ii)

23.
(i)
(iii)
Sol.
2 +  2
 
+
 
 
+
 
 
+
 
=
......(i)
......(ii)
.....[From (i) and (ii)]
..... (iii)
= 13
= ?
 2 + 2
=

=
13
6
If  and  are the roots of the equation 4x2 – 5x + 2 = 0 find the equation
whose roots are.


 + 3 and 3 + 
(ii)  and

1
2
1
2
and
(iv)  +
and  + 
(5 marks)



2
4x – 5x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = 4, b = – 5, c = 2
 and  are the roots of given quadratic equation
–b
– (– 5)
5
Sum of roots =  +  =
=
=
a
4
4
c
2
1
Product of roots =  .  =
=
=
a
4
2
5
1
Now we have =  +  =
and  .  =
4
2
(i)  + 3 and 3 +  are the roots of quadratic equation
Sum of roots =  + 3 + 3 + 
= 4 + 4
= 4 ( + )
5

5
    
= 4×

4
4

= 5
= ( + )2 – 2.
 2 +  2
2
=
=
=
 2 +  2
S C H O O L S E C TI O N
=
1
5
  ×2×
4
2
 
25
–1
16
25 – 16
16
9
16
109
MT
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Product of roots
= ( + 3) × (3 + )
= 32 +  + 9 + 32
= 32 + 32 + 10   
= 3 (2 + 2) + 10 ×
= 3×
1
2
9
+5
16
=
27
5
16
=
27  80
16
=
107
16
1

     2 


9

2
2
     16 


We know that,
x2 – (Sum of roots) x + Product of roots = 0
107
=0
16
Multiplying throughout by 16 we get,
16x2 – 80x + 107 = 0
 x2 – 5x +
 The required equation is 16x2 – 80x + 107 = 0
(ii)


and
are the roots of quadratic equation


 

Sum of roots =


=
=
=
2  2

9
16
1
2
9
1

2
2
     16 and  .   2 


9
2

16 1
9
8
 

Product of roots =
 
= 1
We know that,
x2 – (Sum of roots)x + Product of roots = 0
=
9
x+1=0
8
Multiplying throughout by 8 we get,
8x2 – 9x + 8 = 0
 x2 –
 The required quadratic equation is 8x2 – 9x + 8 = 0
110
S C H O O L S E C TI O N
MT
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2
2
(iii)
and
are the roots of quadratic equation


3

 + 3 = ( + )3 – 3  .  ( + )
125 15
–
=
64
8
125 – 120
=
64
5

3 +  3 =
64
2  2

Sum of roots =


=
=
=
=
Product of roots =
=
=
3  3

5
64
1
2
5
2

64 1
5
32
2  2



. 
1
2
5
1

2
2
     64 and  .   2 


1

  .   2 


We know that,
x2 – (Sum of roots) x + Product of roots = 0
5
1
 x2 –
x+
=0
32
2
Multiplying throughout by 32 we get,
 The required quadratic equation is 32x2 – 5x + 16 = 0
1
1
and  +
are the roots of quadratic equation


1
1
 
Sum of roots =  


1 1

= 
 

=  
(iv)  +

=  
5
5
 4
= 4
1
2
5 5 2
  
=
4 4 1
S C H O O L S E C TI O N
5
1

     4 and  .   2 


111
MT
ALGEBRA
=
=
Product of roots =
=
=
=
=
=
=
=
EDUCARE LTD.
5 10

4
4
15
4
1 
1

       
 


 
1
. 


  .
2  2
1
. 

.
.
9
1
1
1
9

2
2
 16 
1
1
     2 and     16 
2
2
2
1  9
2 
2

   1  
2  16 1  
1
1 9 2
 
2 8 1
4  9  16
8
29
8
We know that,
x2 – (Sum of roots) x + Product of roots = 0
15
29
 x2 –
x+
=0
4
8
Multiplying throughout by 8 we get,
8x2 – 30x + 29 = 0
 The required quadratic equation 8x2 – 30x + 29 = 0.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
17.
Sol.
If  and  are the roots of equation ax2 + bx + c = 0. find the value of
α
β
+
(3 marks)
β
α.
ax2 + bx + c = 0
 and  are the roots of given quadratic equation.
We know that,
 +  =
c
–b
and . =
a
a

β
Now,  + 
=
 2 + 2
.
2
=
( +  ) – 2.
.
2
=
112
c
 –b

 – 2×
a
 a 
c
a
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
=
=
=

α β
+
β α =
 b2 2c  c
 2 –
÷
a  a
a
b2 – 2ac a
×
a2
c
b2 – 2ac
ac
b2  2ac
ac
EXERCISE - 2.7 (TEXT BOOK PAGE NO. 54) :
7.
Sol.
If one root of the quadratic equation kx2 – 20x + 34 = 0 is 5 – 2 2 , find k.
(3 marks)
kx2 – 20x + 34 = 0
Comparing with ax2 + bx + c = 0 we have a = k, b = – 20, c = 34
Let  and  be the roots of given quadratic equation.

= 5–2 2


= 5+2 2

 + 
=
– (– 20)
–b
20
=
=
a
k
k
20
k
20
=
k
20
=
10
 5–2 2 +5+2 2 =

10

k

k
[Given]
[  = 5 – 2 2 and  = 5 + 2 2 ]
= 2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :
15.
Sol.
If one root of the equation x2 – 10x + 2k = 0 is 5 – 3 , find k. (3 marks)
x2 – 10x + 2k = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = – 10, c = 2k
Let  and  be the roots of given quadratic equation.

= 5–
3


= 5+
3

. 

(5 –
c
2k
=
= 2k
a
1
3)
= 2k



(5)2 – ( 3 )2
25 – 3
22
= 2k
= 2k
= 2k

22
2

k
S C H O O L S E C TI O N
3 ) (5 +
=
[Given]
[  = 5 –
3 and  = 5 +
3]
= k
= 11
113
MT
ALGEBRA
16.
Sol.
EDUCARE LTD.
If one root of the quadratic equation x2 + 6x + k = 0 is h + 2 6 , find h and k.
(3 marks)
2
x + 6x + k = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 6, c = k
Let  and  be the roots of given quadratic equation.




 + 
= h+2 6
= h–2 6
=
 h+2 6 +h–2 6 =

2h =

h
=

h
=

. 
=
 (h + 2 6 ) (h – 2 6 )=

(h)2 – (2 6 )2 =

(– 3)2 – 4 × 6 =

9 – 24 =

k =
[Given]
–b
–6
=
=–6
a
1
–6
[  = h + 2 6 and  = h – 2 6 ]
–6
–6
2
–3
c
k
=
=k
a
1
k
k
k
k
– 15
[  = h + 2 6 and  = h – 2 6 ]
[ h = – 3]
 h = – 3 and k = – 15
o
To form quadratic equation if its roots are given :
If  and  are the roots of the quadratic equation in variable x then,
x= 
and
x=
 x–=0
and
x–=0
 (x – ) (x – ) = 0
 x2 – ( + ) x +  = 0
i.e. x2 – (Sum of the roots) x + Product of the roots = 0
EXERCISE - 2.8 (TEXT BOOK PAGE NO. 56) :
1.
(i)
Sol.
Form the quadratic equation if its roots are
5 and – 7
The roots of the quadratic equation are 5 and – 7.
Let  = 5 and  = 7
  +  = 5 + (– 7) = 5 – 7 = – 2
and . = 5 × – 7 = – 35
We know that,
x2 – ( + )x + . = 0
 x2 – (– 2)x + (– 35) = 0

x2 + 2x – 35
= 0
(2 marks)
 The required quadratic equation is x2 + 2x – 35 = 0
(ii)
Sol.
1
–3
and
2
4
The roots of the quadratic equation are
Let  =
114
(2 marks)
1
–3
and  =
2
4
1
–3
and
2
4
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
 3 
1
2–3
–1
+ 
=
 =
2
4
4
 4 
1
–3
–3
and . =
×
=
2
4
8
We know that,
x2 – ( + )x + a.b = 0
 +=
 1 
 3 
 x2 –   x + 
 = 0
 4
 8 
x
3

x2 +
–
= 0
4
8
Multiplying throughout by 8 we get,
8x2 + 2x – 3 = 0
 The required quadratic equation is 8x2 + 2x – 3 = 0
(iii)
Sol.
– 3 and – 11
The roots of the quadratic equation are -3 and -11.
Let  = – 3 and  = – 11
  +  = – 3 + (– 11) = – 3 – 11 = – 14
and . = – 3 × – 11 = 33
We know that,
x2 – ( + )x + .= 0
 x2 – (– 14)x + 33 = 0
 x2 + 14x + 33 = 0
(2 marks)
 The required quadratic equation is x2 + 14x + 33 = 0
(iv)
– 2 and
Sol.
11
2
(2 marks)
The roots of the quadratic equation are – 2 and
Let  = – 2 and 

 + 
and .
=
11
2
11
.
2
11
– 4 + 11
7
=
=
2
2
2
11
= –2×
= – 11
2
= –2+
We know that,
x2 – ( + )x + . = 0
7

x2 – x + (– 11) = 0
2
7

x2 – x – 11 = 0
2
Multiplying throughout by 2 we get,
2x2 – 7x – 22 = 0
 The required quadratic equation is 2x2 – 7x – 22 = 0
(v)
Sol.
1
–1
and
2
2
The roots of the quadratic equation are
S C H O O L S E C TI O N
(3 marks)
1
–1
and
.
2
2
115
MT
ALGEBRA
1
and 
2
 1 
1
 +=
+  
2
 2
1
1
and . =
×
2
2
We know that,
x2 – ( + )x + .
Let  =
EDUCARE LTD.
–1
2
1
1
– =0
=
2
2
-1
=
4
=
= 0
 1 
x2 – 0x +   = 0
 4
1

x2 –
= 0
4
Multiplying throughout by 4 we get,
4x2 – 1 = 0

 The required quadratic equation is 4x2 – 1 = 0
(vi)
Sol.
0 and – 4
The roots of the quadratic equation are 0 and – 4.
Let  = 0 and  = – 4
  +  = 0 + (– 4) = – 4
and . = 0 × – 4 = 0
We know that,
x2 – ( + )x + . = 0
 x2 – (– 4)x + 0 = 0
 x2 + 4x = 0
(2 marks)
 The required quadratic equation is x2 + 4x = 0
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
21.
(i)
Sol.
From the quadratic equation whose roots are.
3 and 10
The roots of the quadratic equation are 3 and 10
Let  = 3 and  = 10
  +  = 3 + 10 = 13
 .  = 3 × 10 = 30
We know that,
x2 – ( + ) x + . = 0
 x2 – 13x + 30 = 0
(2 marks)
 The required quadratic equation is x2 – 13x + 30 = 0
(ii)
Sol.
3
–2
and
4
3
The roots of the quadratic equation are
(3 marks)
3
–2
and
4
3
3
–2
and  =
4
3
3  –2
3 2
9–8
1

–
 +  =
=
=
 =
4  3 
4 3
12
2
Let  =
116
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
3 –2
–1

=
4
3
2
We know that,
x2 – ( + ) x +  . 
. =
 1
1
x + – 
12
 2
1
1

x2 –
x–
12
2
Multiplying throughout by 12
12x2 – x – 6

x2 –
= 0
= 0
= 0
we get,
= 0
 The required quadratic equation is 12x2 – x – 6 = 0
(iii)
Sol.
– 5 and 9
The roots of the quadratic equation are – 5 and 9
Let  = – 5 and  = 9
 +=–5+9=4
 .  = – 5 × 9 = – 45
We know that,
x2 – ( + ) x +  .  = 0

x2 – 4x + (– 45) = 0

x2 – 4x – 45 = 0
(2 marks)
 The required quadratic equation is x2 – 4x – 45 = 0
(iv)
– 3 and
Sol.
5
2
(3 marks)
The roots of the quadratic equation are – 3 and
Let  = – 3 and  =
 +=–3+
5
2
5
2
5
–6  5
–1
=
=
2
2
2
5
–15
=
2
2
We know that,
x2 – ( + ) x + . = 0
. = – 3 ×

 –1
15
x2 –   x –
= 0
2
2
Multiplying throughout by 2 we get,
2x2 + x – 15 = 0
 The required quadratic equation is 2x2 + x – 15 = 0
(v)
Sol.
3
–3
and
4
4
(3 marks)
The roots of the quadratic equation are
Let  =
S C H O O L S E C TI O N
3
–3
and
4
4
3
–3
and  =
4
4
117
MT
ALGEBRA
3  –3


4  4 
3 –3

.=
=
4
4
We know that,
x2 – ( +
 +=
EDUCARE LTD.
3 3
–
=0
4 4
–9
16
=
) x +  . 
=
0
 –9
x2 – (0) x + 
= 0

 16 
9

x2 – 0 –
= 0
16
9

x2 –
= 0
16
Multiplying throughout by 16 we get,
16x2 – 9
= 0

 The required quadratic equation is 16x2 – 9 = 0
(vi)
Sol.
0 and – 6
The roots of the quadratic equation are 0 and – 6
Let  = 0 and  = – 6
  +  = 0 + (– 6) = 0 – 6 = – 6
.=0×–6=0
We know that,
x2 – ( + ) x +  . 
= 0

x2 – (– 6) x + 0
= 0

x2 + 6x
=
0
(2 marks)
 The required quadratic equation is x2 + 6x = 0.
EXERCISE - 2.8 (TEXT BOOK PAGE NO. 56) :
2.
(i)
Sol.
Form the quadratic equation if its one of the root is
(3 marks)
3–2 5
If one of the root of the quadratic equation is 3 – 2 5 ,
then the other root is 3 + 2 5
  = 3 – 2 5 and  = 3 + 2 5
 + = 3–2 5 + 3+2 5 =6
and . =
3 – 2 5  × 3 + 2 5 
(3) – 2 5 
2
2
=
=
9–4×5
=
9 – 20
=
– 11
We know that,
x2 – ( + )x + .

x2 – 6x + (– 11)

x2 – 6x – 11
= 0
= 0
= 0
 The required quadratic equation is x2 – 6x – 11 = 0
(ii)
Sol.
118
4–3 2
If one of the root of the quadratic equation is 4 – 3 2 ,
then the other root is 4 – 3 2
(3 marks)
S C HO O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
  = 4 – 3 2 and  = 4 + 3 2
  +  = 4–3 2 + 4+3 2 =8
and .
=
4 – 3 2  × 4 + 3 2 
(4) – 3 2 
2
=
= 16 – 9 × 2
= 16 – 18
= –2
2
We know that,
x2 – ( + )x + . = 0
 x2 – 8x + (– 2)
= 0
 x2 – 8x – 2
= 0
 The required quadratic equation is x2 – 8x – 2 = 0
(iii)
Sol.
2+ 3
If one of the root of the quadratic equation is
then the other root is 2 – 3
Let  = 2 + 3 and  = 2 – 3
 +=
2
3 
and .
2 –
=
3 = 2 2
 2  3 × 
 2 –  3 
2
2 –
3
(3 marks)
2+ 3,

2
=
= 2–3
= –1
We know that,
x2 – ( + )x + . = 0
 x2 – 2 2 x   –1
= 0

= 0
x2 – 2 2 x – 1
 The required quadratic equation is x2 – 2 2 x – 1 = 0
(iv)
Sol.
(3 marks)
2 3 –4
If one of the root of the quadratic equation is 2 3 – 4 ,
then the other root is 2 3 + 4
Let  = 2 3 – 4 and  = 2 3 + 4
  +  = 2 3 – 4+2 3+4 = 4 3
and .
=
 2 3 – 4 +  2
2 3  – (4)
3+4

2
2
=
= (4 × 3) – 16
= 12 – 16
= –4
We know that,
x2 – ( + )x + . = 0
 x2 – 4 3 x + (– 4) = 0
 x2 – 4 3 x – 4
= 0
 The required quadratic equation is x2 – 4 3 x – 4 = 0.
S C H O O L S E C TI O N
119
MT
ALGEBRA
(v)
(3 marks)
2+ 5
Sol.
EDUCARE LTD.
If one of the root of the quadratic equation is 2 + 5 , then the other
root is 2 – 5 .
Let  = 2 + 5 and  = 2 – 5
  +  = 2+ 5 +2–
and .
5 =4
=
 2 + 5  × 2 – 5 
(2) –  5 
2
=
= 4–5
= –1
2
We know that,
x2 – ( + )x + . = 0
 x2 – 4x + (-1)
= 0
 x2 – 4x – 1
= 0
 The required quadratic equation is x2 – 4x – 1 = 0
(vi)
Sol.
5 – 3
If one of the root of the quadratic equation is
root is 5 + 3 .
Let  = 5 – 3 and  = 5 + 3
 +=
5 –
3 +
and .
(3 marks)
5 –
3 , then the other
5+ 3 = 2 5
=
=
 5 – 3 ×  5 + 3
 5 – ( 3 )
2
2
= 5–3
= 2
We know that,
x2 – ( + )x + . = 0

x2 – 2 5 x + 2
= 0
 The required quadratic equation is x2 – 2 5 x – 2 = 0
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
22.
Form the quadratic equation whose one of the root is.
(i)
1–3 5
Sol.
(3 marks)
The one of the root of the quadratic equation is 1 – 3 5
then the other root is 1 + 3 5
  = 1 – 3 5 and  = 1 + 3 5
  + 
 . 
= 1 – 3 5 +1+ 3 5 = 2
=
=
1 – 3 5   1  3 5 
(1) –  3 5 
2
2
= 1–9×5
= 1 – 45
= – 44
120
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
We know that,
x2 – ( + ) x + . = 0
 x2 – 2x + (– 44)
= 0
 x2 – 2x – 44
= 0
 The required quadratic equation is x2 – 2x – 44 = 0.
(ii)
(3 marks)
3–2 3
Sol.
If one of the root of the quadratic equation is 3 – 2 3 then the other root
is 3 + 2 3
Let  = 3 – 2 3 and  = 3 + 2 3

 +  =3 – 2 3 + 3 + 2 3 = 6
 . 
=
=
 3 – 2 3  × 3 + 2 3 
(3) –  2 3 
2
2
= 9 – 12
= –3
We know that,
x2 – ( + ) x + . = 0
 x2 – 6x + (– 3)
= 0
 x2 – 6x – 3
= 0
 The required quadratic equation is x2 – 6x – 3 = 0
(iii)
Sol.
3 – 7
If one of the root of the quadratic equation is
is
3 –
7 then the other root
3+ 7
Let  =

(3 marks)
3 –
7 and  =
3+ 7
=
7 +
 + 
3 –
3+ 7
= 2 3
 . 
=
=
 3 – 7  3 + 7 
 3 –  7 
2
2
= 3–7
= –4
We know that,
x2 – ( + ) x + . = 0
 x2 – 2 3 + (– 4)
= 0
 x – 2 3 – 4
= 0
2
 The required quadratic equation is x2 – 6x – 3 = 0
(iv)
Sol.
4 2 –3
(3 marks)
If one of the root of the quadratic equation is 4 2 – 3 then the other root
is 4 2 + 3
Let  = 4 2 – 3 and  = 4 2 + 3
S C H O O L S E C TI O N
121
MT
ALGEBRA

 + 
= 4 2 –3 + 4 2+3
= 8 2
. 
=
 4 2 – 3 ×  4
 4 2  – (3)
2
=
= (16 × 2) – 9
= 32 – 9
= 23
2+3
EDUCARE LTD.

2
We know that,
x2 – ( + ) x + . = 0
 x2 – 8 2 x + 23
= 0
 The required quadratic equation is x2 – 8 2 x + 23 = 0
(v)
Sol.
(3 marks)
2+3 5
If one of the root of the quadratic equation is 2 + 3 5 then the other root
is 2 – 3 5
Let  = 2 + 3 5 and  = 2 – 3 5

 + 
= 2+3 5 + 2– 3 5
= 4
 . 
=
 2 + 3 5  2 – 3 5 
(2) – 3 5 
2
=
= 4 – (9 × 5)
= 4 – 45
= – 41
2
We know that,
x2 – ( + ) x + . = 0
 x2 – 4x + (– 41)
= 0
 x2 – 4x – 41
= 0
 The required quadratic equation is x2 – 4x – 41 = 0.
(vi)
7 –
Sol.
(3 marks)
2
The one of the root of the quadratic equation is
root is
 =

7 –
2 then the other
7+ 2
7 –
2 and  =
 + 
=
 . 
=
=
7+ 2
7 –
2 +
7+ 2 = 2 7
 7 – 2  7 + 2 
 7  –  2
2
2
= 7– 2
= 5
We know that,
x2 – ( + ) x + . = 0
 x2 – 2 7 + 5
= 0
 The required quadratic equation is x2 – 2x – 44 = 0.
122
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
EXERCISE - 2.8 (TEXT BOOK PAGE NO. 56) :
3.
Sol.
If the sum of the roots of the quadratic is 3 and sum of their cubes is
63, find the quadratic equation.
(4 marks)
Let  and  be the roots of a quadratic equation.
 +=3
[Given]
and3 + 3 = 63
We know that,
.......(i)
x2 – ( + )x + . = 0
3
3
3
Also,  +  = ( + ) – 3 . ( + )

63 = (3)3 – 3 . (3) [  +  = 3 and 3 + 3 = 63]

63 = 27 – 9 .

9 . = 27 – 63

9 . = – 36

. 
=




. 
x – ( + )x + .
x2 – 3x + (– 4)
x2 – 3x – 4
=
=
=
=
2
– 36
9
–4
0
0
0
[From (i)]
[  +  = 3 and . = – 4]
 The required quadratic equation is x2 – 3x – 4 = 0.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
25.
Sol.
If  +  = 5 and 3 + 3 = 35, find a quadratic equation whose roots are 
and .
(4 marks)
 and  are the roots of a quadratic equation
 +=5
[Given]
3 + 3 = 35
We know that,
x2 – ( + ) x +  .  = 0
......(i)
3
3
3
Also,  +  = ( + ) – 3  .  ( + )

35 = (5)3 – 3 . (5) [  +  = 5 and 3 + 3 = 35]

35 = 125 – 15 .
15 . = 125 – 35


15 . = 90
90

 .  =
15

 .  = 6

x2 – ( + ) x + . = 0
[From (i)]
2

x – 5x + 6 = 0
 The required quadratic equation is x2 – 5x + 6 = 0.
4.
Sol.
If the difference of the roots of the quadratic equation is 5 and the
difference of their cubes is 215, find the quadratic equation. (4 marks)
Let  and  be the roots of a quadratic equation.
  –  = 5 and3 – 3 = 215
[Given]
We know that,
.......(i)
x2 – ( + )x + . = 0
S C H O O L S E C TI O N
123
MT
ALGEBRA




Also, 3 – 3
215
215
215 – 125
90

. 

. 
Now,  – 2
(5) 2
25
25 + 24
( + )2
Taking square root on
 + 
x2 – ( + )x + .
x2 – (7)x + 6 = 0
x2 – 7x + 6 = 0







=
=
=
=
=
EDUCARE LTD.
( – )3 + 3 . ( – )
(5)3 + 3 . (5) [ + = 5 and 3 – 3 = 215]
125 + 15 .
15 .
15 .
90
15
= 6
= ( + )2 – 4.
= ( + )2 – 4 (6) [ . = 5 and . = 6]
= ( + )2 – 24
= ( + )2
= 49
both the sides, we get;
= +7
= 0
[From (i)]
2
or
x – (– 7)x + 6 = 0
or
x2 + 7x + 6 = 0
=
 The required quadratic equation is x2 – 7x + 6 = 0 or x2 + 7x + 6 = 0.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
24.
Sol.
If the difference of the roots of the quadratic equation is 3 and difference
between their cubes is 189, find the quadratic equation.
(4 marks)
Let  and  be the roots of a quadratic equation
  –  = 3 and 3 – 3 = 189
[Given]
We know that,
x2 – ( + ) x + . = 0
......(i)
Also, 3 – 3 = ( – )3 + 3. ( – )

189 = (3)3 + 3. (3) [ . = 3 and 3 – 3 = 189]

189 = 27 + 9 .
189 – 27 = 9 .


162 = 9 .









162
9
. 
Now, ( – )2
(3) 2
9
9 + 72
( + )2
Taking square root on
 + 
x2 – ( +) x + .
x2 – (9)x + 18 = 0
x2 – 9x + 18 = 0
= . 
= 18
= ( + )2 – 4.
= ( + )2 – 4 (18) [  –  = 3 and . = 18]
= ( + )2 – 72
= ( + )2
= 81
both the sides we get,
= +9
= 0
[From (i)]
2
or
x – (– 9)x + 18 = 0
or
x2 + 9x + 18 = 0
 The required quadratic equation is x2 – 9x + 18 = 0 or x2 + 9x + 18 = 0.
124
S C H O O L S E C TI O N
MT
26.
Sol.
ALGEBRA
EDUCARE LTD.
If the difference of the roots of a quadratic equation is 4 and the
difference of their cubes is 208, find the quadratic equation. (4 marks)
Let  and  be the roots of a quadratic equation
  –  = 4 and 3 – 3 = 208
[Given]
We know that,
x2 – ( + ) x + . = 0
......(i)
Also, 3 – 3 = ( – )3 + 3. (.)

208 = (4)3 + 3. (4)
[. =4 and3–3= 208]

208 = 64 + 12 .

208 – 64 = 12 .

144 = 12 .
144
= . 

12

.  = 12
Now, ( – )2 = ( + )2 – 4.

(4) 2 = ( + )2 – 4 (12)
[ – = 4 and . = 12]

16 = ( + )2 – 48

16 + 48 = ( + )2

( + )2 = 64
Taking square root on both the sides we get,
 +  = +8

x2 – ( + )x + . = 0
[From (i)]

x2 – (8)x + 12 = 0 or
x2 – (– 8)x + 12 = 0

x2 – 8x + 12 = 0 or
x2 + 8x + 12 = 0
 The required quadratic equation is x2 – 8x + 12 = 0 or x2 + 8x + 12 = 0
27.
If one root of the quadratic equation ax 2 + bx + c = 0 is the square of the
other, show that b3 + a2c + ac2 = 3 abc.
(4 marks)
To prove : b3 + a2c + ac2 = 3abc
Proof :
ax2 + bx + c = 0
[Given]
Let  and  be the roots of given quadratic equation

= 2
[Given]
–b
c
  + 
=
and. =
a
a
c
2
  .
=
[  = 2]
a
c
 3
=
a
We know that,
( + )3 = 3 + 3 + 3 . ( + )
3

 –b


 a 

– b3
a3
 –b
= (2)3 + 3 + 3 2. 

 a 
 – b
= (3)2 + 3 + 33 

a 

– b3
a3
c
c b
 c
=   + – 3× ×
 a
a
a a
2
–b

2
     a and    


 3 c
   a 


c2
c 3bc
– b3

= 2 + – 2
3
a
a
a
a
Multiplying throughout by a3 we get,
– b3 = ac2 + a2c – 3abc
 3abc = b3 + ac2 + a2c
 b3 + ac2 + a2c = 3abc
Hence proved.
S C H O O L S E C TI O N
125
MT
ALGEBRA
28.
1
1
1
if the sum of roots of the quadratic equation x + p + x + q =
is zero.
r
 p2 + q 2 
.
(5 marks)
Show that the product of the roots is – 
2


Proof :








1
1
+
x+p x+q
=
1
r
x+q+x+p
(x + p) (x + q)
=
1
r
[Given]
2x + p + q
1
=
x + qx + px + pq
r
r (2x + p + q) = x2 + qx + px + pq
2rx + rp + rq = x2 + qx + px + pq
0 = x2 + qx + px – 2rx + pq – pr – qr
2
x + qx + px – 2rx + pq – pr – qr = 0
x2 + x (q + p – 2r) + pq – pr – qr = 0
x2 + (p + q – 2r) x + (pq – pr – qr) = 0
......(i)
Comparing with ax2 + bx + c = 0 we have a = 1, b = p + q – 2r,
c = pq – pr – qr
Let  and  be the roots of equation (i),
2
– (p + q – 2r)
–b
=
= – p – q + 2r
1
a
= 0
[Given]
= 0
[ a + b = – p – q + 2r]
= p+q

 + 



 + 
– p – q + 2r
2r

r
=
Also, .
=

. 
=
pq
2
pq – pr – qr
c
=
= pq – pr – qr
1
a
= pq – pr – qr
= pq – r (p + q)
 p + q
= pq – 
 (p + q)
2 
= pq –

. 
Hence proved.
126
EDUCARE LTD.
p + q

 r = 2 
(p  q)2
2
=
2pq – (p + q)2
2
=
2pq – (p2  2pq  q 2 )
2
=
2pq – p2 – 2pq – q 2
2
=
– p2 – q 2
2
=
– p2 – q 2
2
S C H O O L S E C TI O N
MT
o
ALGEBRA
EDUCARE LTD.
Equations reducible to quadratic form :
Sometimes given equation may not be a quadratic equation. But it can be
reduced to quadratic equation by proper substitution. Then by solving it by
any method, we get the values of new variable.
On resubstituting the values we get the value of the old variable.
EXERCISE - 2.9 (TEXT BOOK PAGE NO. 60) :
(i)
Sol.
Solve the following equations :
x4 – 3x2 + 2 = 0
x4 – 3x2 + 2 = 0
2 2

(x ) – 3x2 + 2 = 0
Substituting x2 = m we get,
m2 – 3m + 2 = 0
2

m – 2m – m + 2 = 0
 m (m – 2) – 1 (m – 2) = 0

(m – 2) (m – 1) = 0

m–2=0
or
m–1=0

m=2
or
m=1
Resubstituting m = x2 we get,
x2 = 2
or
x2 = 1
Taking square roots throughout,
x= ± 2
or
(3 marks)
x = +1
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
29.
(i)
Sol.
Solve the following equations :
x4 – 29x2 + 100 = 0
x4 – 29x2 + 100 = 0
2 2

(x ) – 29x2 + 100 = 0
Substituting x2 = m we get,
m2 – 29m + 100 = 0
2
 m – 4m – 25m + 100 = 0
 m(m – 4) – 25(m – 4) = 0

m–4=0
or
m – 25 = 0

m=4
or
m = 25
Resubstituting m = x2 we get,
x2 = 4 or x2 = 25
Taking square root on both the sides, we get,
x=±2
(ii)
Sol.
or
7y4 – 25y2 + 12 = 0
7y4 – 25y2 + 12 = 0

7(y2)2 – 25y2 + 12 = 0
Substituting y2 = m we get,
7m2 – 25m + 12 = 0
2
 7m – 21m – 4m + 12 = 0
 7m(m – 3) – 4(m – 3) = 0

(m – 3)(7m – 4) = 0

m – 3 = 0 or
m = 3 or

S C H O O L S E C TI O N
m = 3 or
(3 marks)
x=±5
(3 marks)
7m – 4 = 0
7m = 4
4
m=
7
127
MT
ALGEBRA
EDUCARE LTD.
Resubstituting m = y2 we get,
4
7
Taking square root on both the sides, we get,
y2 = 3
y=±
3
or
y2 =
or
y=±
2
7
EXERCISE - 2.9 (TEXT BOOK PAGE NO. 60) :
(vii)
Sol.
Solve the following equations :
3x4 – 13x2 + 10 = 0
3x4 – 13x2 + 10 = 0

3 (x2)2 – 13x2 + 10 = 0
Substituting x2 = m we get,
3m2 – 13m + 10 = 0
2
 3m – 3m – 10m + 10 = 0
 3m (m – 1) – 10 (m – 1) = 0

(m – 1) (3m – 10) = 0

m–1=0
or

m=1
or

m=1
or
Resubstituting m = x2 we get,
(3 marks)
3m – 10 = 0
3m = 10
10
m=
3
10
3
Taking square root on both the sides we get,
(v)
x2 +
Sol.
x2 = 1
or
x2 =
x = +1
or
x= ±
12
=7
x2
(3 marks)
12
= 7
x2
Multiplying throughout by x2 we get,
x4 + 12 = 7x2
2 2
(x ) – 7x2 + 12 = 0
Substituting x2 = m we get,
m2 – 7m + 12 = 0
2
m – 4m – 3m + 12 = 0
m (m – 4) – 3 (m – 4) = 0
(m – 4) (m – 3) = 0
m–4=0
or
m–3=0
m=4
or
m=3
Resubstituting m = x2 we get,
x2 = 4
or
x2 = 3
Taking square root on both the sides we get,
x2 +






x=+2
(iv)
Sol.
128
10
3
12
35y2 + y2 = 44
12
35y2 + y 2 = 44
or
x= ± 3
(3 marks)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.







Multiplying throughout by y2 we
35y4 + 12 = 44y2
2 2
35 (y ) – 44y2 + 12 = 0
Substituting y2 = m we get,
35m2 – 44m + 12 = 0
2
35m – 14m – 30m + 12 = 0
7m (5m – 2) – 6 (5m – 2) = 0
(5m – 2) (7m – 6) = 0
5m – 2 = 0
or
5m = 2
or
2
m=
or
5
Resubstituting m = y2, we get;
y= ±
y=
get,
7m – 6 = 0
7m = 6
6
m=
7
2
5
or
y= ±
2
5
or
y=
6
7
6
7
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
29.
(iii)
Sol.
Solve the following equations :
2
=7
6m2 +
m2
2
6m2 +
= 7
m2
Multiplying throughout by m2, we get
6m4 + 2 = 7m 2
2 2

6(m ) – 7m2 + 2 = 0
Substituting m2 = x we get,

6x2 – 7x + 2 = 0
2

6x – 3x – 4x + 2 = 0
 3x (2x – 1) – 2 (2x – 1) = 0

(2x – 1)(3x – 2) = 0

2x – 1 = 0
or
3x – 2 = 0

2x = 1
or
3x = 2
1
2

x=
or
x=
2
3
Resubstituting x = m2 we get,
1
2
or
m2 =
m2 =
2
3
Taking square root on both the sides we get,
m=±
1
2
or
m=±
(3 marks)
2
3
EXERCISE - 2.9 (TEXT BOOK PAGE NO. 60) :
(viii)
Sol.
Solve the following equations :
15
2y2 + y2 = 12
(3 marks)
15
2y2 + y 2 = 12
S C H O O L S E C TI O N
129
MT
ALGEBRA
EDUCARE LTD.
Multiplying throughout by y2 we get,
2y4 + 15 = 12y2
2 2

2 (y ) – 12y2 + 15 = 0
Substituting y2 = m we get,
2m2 – 12m + 15 = 0
Comparing with am2 + bm + c = 0 we a = 2, b = – 12, c = 15
b2 – 4ac = (– 12)2 – 4 (2) (15)
= 144 – 120
= 24
m =
=
=
=
=
=

b2 – 4ac
2a
– (–12) ± 24
–b ±
2(2)
– (–12) ± 24
2(2)
12 ± 2 6
4
(
2 6± 6
)
4
6± 6
2
6+ 6
or
2
2
Resubstituting m = y we get,
m=
m=
6– 6
2
6+ 6
6– 6
or
y2 =
2
2
Taking square root on both the sides we get,
y2 =
y= ±
6+ 6
2
or
y= ±
6– 6
2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
29.
(x)
Sol.
130
Solve the following equations :
2 (y2 – 6y)2 – 8 (y2 – 6y + 3) – 40 = 0
2 (y2 – 6y)2 – 8 (y2 – 6y + 3) – 40 = 0
Substituting y2 – 6y = m we get,

2m2 – 8 (m + 3) – 40 = 0

2m2 – 8m – 24 – 40 = 0

2m2 – 8m – 64 = 0
Dividing throughout by 2 we get,
m2 – 4m – 32 = 0
2

m – 8m + 4m – 32 = 0

m (m – 8) + 4 (m – 8) = 0

(m – 8) (m + 4) = 0

m – 8 = 0 or m + 4 = 0

m = 8 or m = – 4
Resubstituting m = y2 – 6y we get,
y2 – 6y = 8 .....(i) or y2 – 6y = – 4
From (i),
y2 – 6y = 8

y2 – 6y – 8 = 0
(3 marks)
........(ii)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
Comparing with ay2 + by + c = 0 we have a = 1, b = – 6, c = – 8
b2 – 4ac = (– 6)2 – 4 (1) (– 8)
= 36 + 32
= 68
y
=
=
=
6 ± 4 × 17
2
=
6 ± 2 17
2
=

y = 3 + 17
y2 – 6y
2

y – 6y + 4
Comparing with ay2 + by
b2 – 4ac
From (ii),
y
17

2
3 ± 17
or
=
=
+c
=
=
=
y = 3 – 17
–4
0
= 0 we have a = 1, b = – 6, c = 4
(– 6)2 – 4 (1) (4)
36 – 16
20
=
=
=
=
=
y= 3+ 5
3 ±
2
=
=

b2 – ac
2a
– (– 6) ± 68
2 (1)
–b±
b2 – 4ac
2a
–b ±
– (– 6) ± 20
2 (1)
6±
4×5
2
6±2 5
2
2 3± 5


2
3± 5
or 3 –
5
 y = 3 + 17 or y = 3 – 17 or y = 3 + 5 or y = 3 –
(ix)
Sol.
(y2 + 5y) (y2 + 5y – 2) – 24 = 0
(y2 + 5y) (y2 + 5y – 2) – 24 = 0
Substituting y2 + 5y = m we get,

m (m – 2) – 24 = 0

m2 – 2m – 24 = 0
2

m – 6m + 4m – 24 = 0
 m (m – 6) + 4 (m – 6) = 0

(m – 6) (m + 4) = 0

m – 6 = 0 or m + 4 = 0

m = 6 or m = – 4
S C H O O L S E C TI O N
5
(3 marks)
131
MT
ALGEBRA












Resubstituting m = y2 +
y2 + 5y = 6 ......(i)
From (i),
y2 + 5y
2
y + 5y – 6
2
y – y + 6y – 6
y (y – 1) + 6 (y – 1)
(y – 1) (y + 6)
y–1=0
y=1
From (ii),
y2 + 5y
y2 + 5y + 4
2
y + 4y + y + 4
y (y + 4) + 1 (y + 4)
(y + 4) (y + 1)
y+4=0
y=–4
5y
or
=
=
=
=
=
or
or
=
=
=
=
=
or
or
EDUCARE LTD.
we get,
y2 + 5y = – 4 .....(ii)
6
0
0
0
0
y+6=0
y=–6
–4
0
0
0
0
y+1=0
y=–1
 y = 1 or y = – 6 or y = – 4 or y = – 1.
(iv)
Sol.








(p2 + p)(p2 + p – 3) = 28
(4 marks)
(p2 + p)(p2 + p – 3) = 28
Substituting p2 + p = m we get,
m (m – 3) = 28
m2 – 3m – 28 = 0
m2 – 7m + 4m – 28 = 0
m (m – 7) + 4 (m – 7) = 0
(m – 7) (m + 4) = 0
m – 7 = 0 or m + 4 = 0
m = 7 or m= –4
Resubstituting m = p2 + p we get,
p2 + p = 7 or p2 + p = – 4
2
p + p – 7= 0...........(i) or p2 + p + 4 = 0
........(ii)
2
From (i),
p +p–7 = 0
Comparing with ap2 + bp + c = 0 we have a =1, b = 1, c = – 7
b2 – 4ac = (1)2 – 4 (1) (–7)
= 1 + 28
= 29
p
=
=
=

132
From (ii),
p2 + p + 4 =
Comparing with ap2 + bp +
b2 – 4ac =
=
=
b2 – 4ac <
–b ±
b2 – 4ac
2a
–1 ± 29
2(1)
–1 ± 29
2
0
c = 0 we have a = 1, b =1, c = 4
(1)2 – 4 (1)(4)
1 – 16
– 15
0
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
 The roots of this equation are not real hence not considered.
p
 p=
=
–1 ± 29
2
–1  29
–1 – 29
or p =
2
2
EXERCISE - 2.9 (TEXT BOOK PAGE NO. 60) :
(ii)
Sol.
Solve the following equations :
(x2 + 2x) (x2 + 2x – 11) + 24 = 0
(x2 + 2x) (x2 + 2x – 11) + 24 = 0
Substituting x2 + 2x = m we get,
m (m – 11) + 24 = 0

m2 – 11m + 24 = 0

m2 – 8m – 3m + 24 = 0

m (m – 8) – 3 (m – 8) = 0

(m – 8) (m – 3) = 0

m – 8 = 0 or m – 3 = 0

m = 8 or m = 3
Resubstituting m = x2 + 2x we get,
or x2 + 2x = 3
x2 + 2x = 8
2
 x + 2x – 8 = 0 ....... (i) or x2 + 2x – 3 = 0 ......(ii)
From (i),
x2 + 2x – 8 = 0

x2 + 4x – 2x – 8 = 0

x(x + 4) – 2 (x + 4) = 0

(x + 4) (x – 2) = 0

x + 4 = 0 or x – 2 = 0

x = – 4 or x = 2
2
From (ii), x + 2x – 3 = 0

x2 + 3x – x – 3 = 0

x (x + 3) – 1(x + 3) = 0

(x + 3) (x – 1) = 0

x + 3 = 0 or x – 1 = 0

x = – 3 or x = 1
(4 marks)
 x = – 4 or x = 2 or x = – 3 or x = 1.
(vi)
Sol.
(x2 + x) (x2 + x – 7) + 10 = 0
(x2 + x) (x2 + x – 7) + 10 = 0
Substituting x2 + x = m we get,
m (m – 7) + 10 = 0

m2 – 7m + 10 = 0

m2 – 5m – 2m + 10 = 0

m (m – 5) – 2 (m – 5) = 0

(m – 5) (m – 2) = 0

m – 5 = 0 or m – 2 = 0

m = 5 or m = 2
Resubstituting m = x2 + x we get,
S C H O O L S E C TI O N
(4 marks)
133
MT
ALGEBRA
x2 + x = 5
 x2 + x – 5 = 0 ....... (i)
From (i),
x2 + x – 5
Comparing with ax2 + bx

b2 – 4ac
or
or
=
+c
=
=
=
x
=

x
=

x






–1  21
2
From (ii),
x2 + x – 2
x2 + 2x – 1x – 2
x (x + 2) – 1(x + 2)
(x + 2) (x – 1)
x+2=0
x=–2
x=
 x=
EDUCARE LTD.
x2 + x = 2
x2 + x – 2 = 0 ......(ii)
0
= 0 we have a = 1, b = 1, c = – 5
(1)2 – 4(1) (– 5)
1 + 20
21
–b
b2 – 4ac
2a
–1  21
2 ×1
–1  21
2
–1 – 21
or x =
2
= 0
= 0
= 0
= 0
or x – 1 = 0
or x = 1
=
–1  21
–1 – 21
or x =
or x = – 2 or x = 1.
2
2
EXERCISE - 2.9 (TEXT BOOK PAGE NO. 60) :
Solve the following equations :
(iii)
1 
1


2  x 2 + 2  - 9  x +  +14 = 0
x 
x


(5 marks)
1 
1


2  x 2 + 2  – 9  x +  + 14 = 0
x 
x


1
Substituting x +
= m
x
Squaring both the sides we get,
Sol.
1

 x + 
x
.......(i)
2
=
m2
 (x)2 + 2 × x ×
=
m2

=
m2
=
m2 – 2
=
=
=
=
0
0
0
0
2





134
1
 1
+  
 x
x
1
x2 + 2 + 2
x
1
x2 + 2
x
Equation (i) becomes
2 (m2 – 2) – 9m + 14
2m2 – 4 – 9m + 14
2m2 – 9m + 10
2
2m 4m – 5m + 10
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
















2m (m – 2) – 5 (m – 2)
(m – 2) (2m – 5)
m–2=0
m=2
=
=
or
or
0
0
2m – 5 = 0
2m = 5
5
m = 2 or m =
2
1
Resubstituting m = x 
we get,
x
1
1
5
= 2 ..... (ii) or x +
=
..... (iii)
x+
x
x
2
1
From (ii), x +
= 2
x
Multiplying throughout by x we get,
x2 + 1 = 2x
2
x – 2x + 1 = 0
(x – 1)2 = 0
Taking square root on both the sides we get,
x–1 = 0
x = 1
1
5
From (iii),
x+
=
x
2
Multiplying throughout by 2x, we get;
2x2 + 2 = 5x
2
2x – 5x + 2 = 0
2x2 – 4x – x + 2 = 0
2x (x – 2) – 1 (x – 2) = 0
(x – 2) (2x – 1) = 0
x – 2 = 0 or 2x – 1 = 0
x = 2 or 2x = 1
1
x = 2 or x =
2
1
x = 1 or x = 2 or x =
2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :
29.
Solve the following equations :
(viii)
1 
1


12  x 2 + 2  – 56  x +  + 89 = 0
x
x 


1 
1


12  x 2 + 2  – 56  x +  + 89 = 0
x
x 


1
Substituting x +
= m
x
Squaring both the sides we get,
Sol.
(5 marks)
.....(i)
2



1

 x + 
x
1
1
x2 + 2 × x ×
+ 2
x
x
1
x2 + 2 + 2
x
1
x2 + 2
x
S C H O O L S E C TI O N
=
m2
=
m2
=
m2
=
m2 – 2
135
MT
ALGEBRA






















Equation (i) becomes,
12 (m2 – 2) – 56m + 89 = 0
12m2 – 24 – 56m + 89 = 0
12m2 – 56m + 65 = 0
2
12m – 30m – 26m + 65 = 0
6m (2m – 5) – 13 (2m – 5) = 0
(2m – 5) (6m – 13) = 0
2m – 5 = 0 or 6m – 13 = 0
2m = 5 or 6m = 13
5
13
m=
or m =
2
6
1
Resubstituting m = x 
we get,
x
1
5
1
13
x +
=
.....(ii) or x +
=
.....(iii)
x
2
x
6
1
5
x
=
From (ii),
x
2
Multiplying throughout by 2x we get,
2x2 + 2 = 5x
2x2 – 5x + 2 = 0
2x2 – 4x – x + 2 = 0
2x (x – 2) – 1 (x – 2) = 0
(x – 2) (2x – 1) = 0
x – 2 = 0 or 2x – 1 = 0
x = 2 or 2x = 1
1
x = 2 or x =
2
1
13
From (iii),
x+
=
x
6
Multiplying throughout by 6x we get,
6x2 + 6 = 13x
2
6x – 13x + 6 = 0
2
6x – 9x – 4x + 6 = 0
3x (2x – 3) – 2 (2x – 3) = 0
(2x – 3) (3x – 2) = 0
2x – 3 = 0 or 3x – 2 = 0
2x = 3 or 3x = 2
3
2
x=
or x =
2
3
 x = 2 or x =
(v)
Sol.
EDUCARE LTD.
1
3
2
or x =
or x =
2
2
3
1

 2 1 
3x + 2  – 4  x –
 –6=0
x 
x


1

 2 1 
3x + 2  – 4 x –  – 6 = 0
x 
x


1
= m
Substituting x –
x
squaring both the sides we get,
(5 marks)
......(i)
2
1

x – 
x

2
1
 1
x2 – 2 . x .
+  
 x
x
136
=
=
m2
m2
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.














1
x2
1
x2 + 2
x
Equation (i) becomes,
3 (m2 + 2) – 4m – 6
3m2 + 6 – 4m – 6
3m2 – 4m
m (3m – 4)
m=0
m=0
x2 – 2 +
=
m2
=
m2 + 2
0
0
0
0
3m - 4 = 0
3m = 4
4
m = 0 or m =
3
1
Resubstituting m = x –
we get,
x
1
1
4
x–
= 0 ........(ii) or x –
=
.......(iii)
x
x
3
1
x–
From (ii),
= 0
x
Multiplying throughout by x, we get;
x2 – 1 = 0
x 2 – 12 = 0
(x + 1)(x – 1) = 0
x + 1 = 0 or x – 1 = 0
x = – 1 or x = 1
1
4
x–
=
From (iii),
x
3
Multiplying throughout by 3x, we get;
3x2 – 3 = 4x
2
3x – 4x – 3 = 0
Comparing with ax2 + bx + c = 0 we have a =3, b = – 4, c = – 3
b2 – 4ac = (– 4)2 – 4 (3)(–3)
= 16 + 36
= 52
x
=
=
=
=
or
or
=
– b + b2 – 4ac
2a
=
– (– 4) ± 52
2(3)
=
4 ± 4×13
6
=
=

x=
2 + 13
3
 x = – 1 or x = 1 or x =
S C H O O L S E C TI O N
4 ± 2 13
6
2 2 ± 13
(
)
6
2 ± 13
=
3
2 – 13
or x =
3
2 + 13
2 – 13
or x =
3
3
137
MT
ALGEBRA
1
 2 1 

9  x + 2  – 3  x –  – 20 = 0
x 
x


1
 2 1

9  x + 2  – 3  x –  – 20 = 0
x 
x 


1
= m
Substituting x –
x
Squaring both the sides we get,
(vi)
Sol.
EDUCARE LTD.
(5 marks)
..........(i)
2



1

x – 
x

1
x2 – 2 + 2
x
1
x2 + 2
x
Equation (i) becomes,
9(m2 + 2) – 3m – 20
9m2 + 18 – 3m – 20
9m2 – 3m – 2
2
9m + 3m – 6m – 2
3m (3m + 1) – 2 (3m + 1)
(3m + 1) (3m – 2)
3m + 1= 0
3m = – 1
1
m= –
3
=
m2
=
m2
=
m2 + 2
0
0
0
0
0
0
3m – 2 = 0
3m = 2
2

or m =
3
1
Resubstituting m = x –
we get,
x
1
1
1
2
x –
= –
..........(ii) or x –
=
...........(iii)
x
3
x
3
1
1
x –
= –
From (ii),
x
3
Multiplying throughout by 3x we get,

3x2 – 3 = – x
2

3x + x – 3 = 0
Comparing with ax2 + bx + c = 0 we have a = 3, b = 1, c = – 3
b2 – 4ac = (1)2 – 4 (3) (– 3)
= 1 + 36
= 37







x
=
=
=
=
=
=
or
or
=
–b ±
b2 – 4ac
2a
=
–1 ± 37
2(3)
=
–1 ± 37
6
1 + 37
–1 – 37
or x =
6
6
1
2
x –
From (iii),
=
x
3
Multiplying throughout by 3x, we get;
3x2 – 3 = 2x
2

3x – 2x – 3 = 0

138
x=
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
Comparing with ax2 + bx + c = 0 we have a = 3, b = –2, c = – 3
b2 – 4ac = (–2)2 – 4 (3)(– 3)
= 4 + 36
= 40
x
=
=
=
=
=

x=
 x=
(vii)
Sol.
– 1 + 10
3
or
b2 – 4ac
2a
–b ±
– 2 ± 40
2(3)
–2±
4 × 10
6
– 2 ± 2 10
6
– 1 ± 10
3
– 1 – 10
3
– 1 + 10
– 1 – 10
1 + 37
–1 – 37
or x =
or x =
or x =
6
6
3
3
1 
1
 2

30  x + 2  – 77  x –  – 12 = 0
x 
x


1
1
 2

30  x + 2  – 77  x –  – 12 = 0
x 
x


1
Substituting x –
= m
x
Squaring both the sides we get,
(5 marks)
.........(i)
2











1

=
x – 
x

1
1
x2 – 2 × x ×
+ 2
=
x
x
1
x2 – 2 + 2
=
x
1
x2 + 2
=
x
Equation (i) becomes,
30 (m2 + 2) – 77m – 12
30m2 + 60 – 77m – 12
30m2 – 77m + 48
2
30m – 45m – 32m + 48
15m (2m – 3) – 16 (2m – 3)
(2m – 3)(15m – 16)
2m = 3
3
m=
2
Resubstituting the m = x –
x –
S C H O O L S E C TI O N
1
3
=
x
2
........(ii)
m2
m2
m2
m2 + 2
=
=
=
=
=
=
or
0
0
0
0
0
0
15m = 16
16
or m =
15
1
we get,
x
or
x –
1
16
=
......(iii)
x
15
139
MT
ALGEBRA
1
3
=
x
2
Multiplying throughout by 2x, we get;
= 3x
2x2 – 2
2
2x – 3x – 2
= 0
2x2 – 4x + x – 2
= 0
2x (x – 2)+ 1 (x – 2)
= 0
(x – 2)(2x + 1)
= 0
x – 2 = 0 or 2x + 1
= 0
x=2
or 2x = – 1
–1
x=2
or x =
2
1
16
x –
From (iii),
=
x
15
Multiplying throughout by 15x we get,
15x2 – 15 = 16x
2
15x – 16x – 15 = 0
15x2 – 25x + 9x – 15 = 0
5x (3x – 5) + 3 (3x – 5) = 0
(3x – 5)(5x + 3) = 0
3x – 5 = 0 or 5x + 3 = 0
3x = 5 or 5x = – 3
5
–3
x=
or x =
3
5
From (ii),














EDUCARE LTD.
 x = 2 or x =
x –
–1
5
–3
or x =
or x =
2
3
5
WORD PROBLEMS
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :
2.
(i)
Sol.
From an equations for the following examples.
The sum of a natural number ‘x’ and its square is 30.
A natural number is ‘x’
 Its square is x2
As per the given condition,
x + x2 = 30
2

x + x - 30 = 0
(ii)
Sol.
The product of two numbers ‘y’ and y – 3 is 42.
Two numbers are ‘y’ and y – 3.
As per the given condition,
y (y – 3) = 42

y2 – 3y – 42 = 0
(iii)
Sol.
140
(1 mark)
(1 mark)
37
The sum of a natural number ‘x’ and its reciprocal is
.
6
A natural number is ‘x’.
1
 It’s reciprocal is
.
x
As per the given condition,
1
37
=
x+
x
6
Multiplying throughout by 6x we get,
6x2 + 6 = 37x
2

6x – 37x + 6 = 0
(1 mark)
S C H O O L S E C TI O N
MT
(iv)
Sol.
(v)
Sol.
EDUCARE LTD.
ALGEBRA
The digit at ten’s place of a two digit number is greater than the square
of digit at unit’s place(x) by 5 and the number formed is 61. (1 mark)
The digit is unit’s place of a two digit number is ‘x’.
 The digit in ten’s place is x2 + 5
As per the given condition,
10 (x2 + 5) + 1 (x) = 61

10x2 + 50 + x – 61 = 0

10x2 + x – 11 = 0
The length of a rectangle (x) is greater than its breadth by 3 cm. The
(1 mark)
area of a rectangle is 70 sq.cm
The length of a rectangle is ‘x’ cm.
 It’s breadth is (x – 3) cm.
As per the given condition,

Area of rectangle = Length × Breadth

70 = x × (x – 3)

70 = x2 – 3x

0 = x2 – 3x – 70

x2 – 3x – 70 = 0
EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :
6.
The sum of a natural number and its reciprocal is
Sol.









10
. Find the number.
3
(3 marks)
Let natural number be ‘x’
1
Its reciprocal is
x
From the given condition,
1
10
=
x+
x
3
Multiplying throughout by 3x,
3x2 + 3 = 10x
3x2 – 10x + 3 = 0
3x2 – 9x – x + 3 = 0
3x (x – 3) – 1 (x – 3) = 0
(3x – 1) (x – 3) = 0
3x – 1 = 0
or x – 3 = 0
3x = 1
or x = 3
1
x=
or x = 3
3
1
because x is natural number
x 
3
x=3
 The natural number is 3.
8.
Sol.
Three times the square of a natural numbers is 363. Find the numbers.
(3 marks)
Let the natural number be ‘x’
From the given condition,
3x2 = 363
363
 x2 =
3
S C H O O L S E C TI O N
141
MT
ALGEBRA
EDUCARE LTD.
 x2 = 121
 x = + 11
[Taking square roots]
x  – 11 because x is a natural number
 x = 11
 The natural number is 11.
11.
Sol.
The sum ‘S’ of the first ‘n’ natural numbers is given by S =
Find ‘n’, if the sum (S) is 276.
n (n + 1)
S=
2
From the given condition,
n (n + 1)
276 =
2
 552 = n2 + n
 n2 + n – 552 = 0
 n2 + 24n – 23n – 552 = 0
 n (n + 24) (n – 23) = 0
 n + 24 = 0
or n – 23 = 0
 n = – 24
or n = 23
n  – 24 because ‘n’ cannot be negative
n (n + 1)
.
2
(3 marks)
 n = 23
WORD PROBLEMS BASED ON NUMBERS
EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :
1.
Sol.
The sum of the squares of two consecutive natural numbers is 113.
Find the numbers.
(3 marks)
Let the two consecutive natural numbers be x and x + 1
As per the given condition,
x2 + (x + 1)2 = 113
2

x + x2 + 2x + 1 = 113
2

2x + 2x + 1 – 113 = 0

2x2 + 2x – 112 = 0
Dividing throughout by 2 we get,
x2 + x – 56 = 0
2

x + 8x – 7x – 56 = 0

x (x + 8) – 7 (x + 8) = 0

(x + 8) (x – 7) = 0

x + 8 = 0 or x – 7 = 0

x = – 8 or x = 7
 x is a natural number
 x  -8
Hence, x = 7
And x + 1 = 7 + 1 = 8
 The two consecutive natural numbers are 7 and 8 respectively.
4.
Sol.
142
The sum of the squares of two consecutive even natural numbers is
(3 marks)
100. Find the numbers.
Let the two consecutive even natural numbers be x and x +2.
As per the given condition.
x2 + (x + 2)2 = 100
2

x + x2 + 4x + 4 = 100
S C H O O L S E C TI O N
MT
EDUCARE LTD.








ALGEBRA
2x2 + 4x + 4 – 100 = 0
2x2 + 4x – 96 = 0
Dividing throughout by 2 we get,
x2 + 2x – 48 = 0
2
x – 6x + 8x – 48 = 0
x (x – 6) + 8 (x – 6) = 0
(x – 6) (x + 8) = 0
x – 6 = 0 or x + 8 = 0
x = 6 or x = – 8
x is an natural number
x –8
Hence x = 6
And x + 2 = 6 + 2 = 8
 The two consecutive even natural numbers are 6 and 8 respectively.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :
30.
(i)
Sol.
(xiii)
Sol.
Solve the following :
Three consecutive odd natural numbers are such that the product of
the first and third is greater than four times the middle by 1. Find the
(3 marks)
numbers.
Let the third consecutive odd natural number be x, x + 2 and x + 4
As per the given condition,
x (x + 4) = 4 (x + 2) + 1

x2 + 4x = 4x + 8 + 1

x2 = 9
Taking square root on both the sides we get,
x = +3
 x is a natural number x  – 3
Hence,
x = 3
x + 2 = 3 + 2 = 5 and x + 4 = 3 + 4 = 7
 The 3 consecutive odd natural numbers are 3, 5 and 7 respectively.
The sum of the squares of five consecutive natural numbers is 1455.
find them.
(4 marks)
Let the five consecutive natural numbers be x, x + 1, x + 2, x + 3 and
x + 4 respectively.
As per the given condition,
x2 + (x + 1)2 + (x + 2)2 + (x + 3)2 + (x + 4)2
= 1455
 x2 + x2 + 2x + 1 + x2 + 4x + 4 + x2 + 6x + 9 + x2 + 8x + 16 – 1455 = 0
 5x2 + 20x + 30 – 1455
= 0
 5x2 + 20x – 1425
= 0
Dividing throughout by 5 we get,
x2 + 4x – 285 = 0
2

x – 15x + 19x – 285 = 0
 x (x – 15) + 19 (x – 15) = 0

(x – 15) (x + 19) = 0

x – 15 = 0 or x + 19 = 0

x = 15 or x = – 19
 x is a natural number x  – 19
Hence x = 15
 x + 1 = 15 + 1 = 16

x + 2 = 15 + 2 = 17
 x + 3 = 15 + 3 = 18

x + 4 = 15 + 4 = 19
 The required five consecutive natural numbers are 15, 16, 17, 18
and 19 respectively.
S C H O O L S E C TI O N
143
MT
ALGEBRA
(x)
Sol.
EDUCARE LTD.
The product of four consecutive positive integers is 840. find the largest
(5 marks)
number.
Let the four consecutive positive integers be x, x + 1, x + 2 and x + 3
As per the given condition,
x × (x + 1) × (x + 2) × (x + 3) = 840

x (x + 3) × (x + 1) (x + 2) = 840
2

(x + 3x) × (x2 + 2x + x + 2) = 840

(x2 + 3x) (x2 + 3x + 2) = 840
Substituting x2 + 3x = m we get,
m (m + 2) = 840

m2 + 2m – 840 = 0

m2 + 30m – 28m – 840 = 0

m (m + 30) – 28 (m + 30) = 0

(m + 30) (m – 28) = 0

m + 30 = 0 or m – 28 = 0

m = – 30 or m = 28
Resubstituting m = x2 + 3x we get,
x2 + 3x = – 30 ......(i) or x2 + 3x = 28 ......(ii)
From (i),
x2 + 3x = – 30
2

x + 3x + 30 = 0
Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = 30
b2 – 4ac = (3)2 – 4 (1) (30)
= 9 – 120
= – 111

b2 – 4ac < 0
 The roots of the above quadratic equation are not real.
Hence not considered.
From (ii),
x2 + 3x = 28
2

x + 3x – 28 = 0

x2 + 7x – 4x – 28 = 0

x (x + 7) – 4 (x + 7) = 0

(x + 7) (x – 4) = 0

x + 7 = 0 or x – 4 = 0

x = – 7 or x = 4
 x is positive integer

x  –7
Hence x = 4
 x+3=4+3=7
 The largest required number is 7.
8
.
75
(4 marks)
(iv)
Divide 40 into two parts such that the sum of their reciprocals is
Sol.
Sum of two parts is 40
Let one of the part is x
 The other part is 40 – x
As per the given condition,
1
1

=
x 40 – x
40 – x + x

=
x (40 – x)


144
40
40x – x 2
40 (75)
=
=
8
75
8
75
8
75
8 (40x – x2)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.








40 × 75
=
8
375 =
x2 – 40x + 375 =
x2 – 25x – 15x + 375 =
x (x – 25) – 15 (x – 25) =
(x – 25) (x – 15) =
x – 25 = 0 or
x = 25 or
If
x = 25 or
then 40 – x = 40 – 25 = 15
40x – x2
40x – x2
0
0
0
0
x – 15 = 0
x = 15
if x = 15
then 40 – x = 40 – 15 = 25
 The two parts are 25 and 15.
(xvi)
Sol.
The difference between two positive integers is 2 and the difference
between their cubes is 56. Find the numbers.
(3 marks)
Let the smaller positive integer be x
 The bigger positive integer is x + 2
As per the given condition,
(x + 2)3 – x3
=
56
 x3 + 3 × (x2) × (2) + 3 × (x) × 22 + 23 – x3 – 56 =
0
 x3 + 6x2 + 12x + 8 – x3 – 56
=
0
2
 6x + 12x – 48
=
0
Dividing throughout by 6 we get,
x2 + 2x – 8 = 0

x2 + 4x – 2x – 8 = 0

x (x + 4) – 2 (x + 4) = 0

(x + 4) (x – 2) = 0

x + 4 = 0 or x – 2 = 0

x = – 4 or x = 2
 x is a positive integer

x  –4
Hence x = 2
 x+2=2+2=4
 The two positive integers are 2 and 4.
(vi)
Sol.
The sum of four times a number and three times its reciprocal is 7. Find
that number.
(3 marks)
Let the number be x
1
 It’s reciprocal is
x
As per the given condition,





1
4 (x) + 3   =
x
3
4x +
=
x
Multiplying throughout by x
4x2 + 3 =
2
4x – 7x + 3 =
4x2 – 4x – 3x + 3 =
4x (x – 1) – 3 (x – 1) =
(x – 1) (4x – 3) =
S C H O O L S E C TI O N
7
7
we get,
7x
0
0
0
0
145
MT
ALGEBRA


x–1=0
x=1

x=1
 The number is 1 or
(viii)
Sol.
EDUCARE LTD.
or 4x – 3 = 0
or 4x = 3
3
or x =
4
3
.
4
A natural number is greater then three times its square root by 4. Find
the number.
(3 marks)
Let the square root of natural number be x
 The natural number = x2
As per the given condition,
x2 = 3x + 4
2

x – 3x – 4 = 0

x2 – 4x + 1x – 4 = 0

(x – 4) (x + 1) = 0

x – 4 = 0 or x + 1 = 0

x = 4 or x = – 1
Square root of natural number cannot be negative

x  –1

x = 4

x2 = 42

x2 = 16
 The natural number is 16.
EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :
5.
Sol.
A natural number is greater than twice its square root by 3. Find the
number.
(3 marks)
Let square root of the natural number be x
 The natural number = x2
As per the given condition,
x2 = 2x + 3
2

x – 2x – 3 = 0

x2 – 3x + 1x – 3 = 0

x (x – 3) + 1 (x – 3) = 0

(x – 3) (x + 1) = 0

x – 3 = 0 or x + 1 = 0

x = 3 or x = – 1
 Square root of a natural number cannot be negative

x  –1

x = 3

x2 = 9
 The natural number is 9.
10.
Sol.
146
A natural number is greater than the other by 5. The sum of their
squares is 73. Find those numbers.
(3 marks)
Let the other natural number be x.
 The first natural number is x + 5.
As per the given condition,
x2 + (x + 5)2 = 73
2
2
 x + x + 10x + 25 – 73 = 0

2x2 + 10x – 48 = 0
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.







Dividing throughout by 2 we get,
x2 + 5x – 24 = 0
2
x – 3x + 8x – 24 = 0
x (x – 3) + 8 (x – 3) = 0
(x – 3) (x + 8) = 0
x – 3 = 0 or x + 8 = 0
x = 3 or x = -8
The number is a natural number
x  –8
Hence x = 3
And x + 5 = 3 + 5 = 8
 The natural numbers are 3 and 8.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 168) :
30.
(xix)
Sol.
Solve the following :
The divisor and quotient of the number 6123 are same and the remainder
is half the divisor. Find the divisor.
(4 marks)
Let divisor of 6123 be ‘x’
 Divisor = Quotient
[Given]
 Quotient = x
x
Remainder =
[Given]
2
We know,
Dividend = Divisor × Quotient + Remainder
x
6123 = x . x +
2
Multiplying throughout by 2,
2 (6123) = 2x2 + x

12246 = 2x2 + x
2

2x + x – 12246 = 0
 2x2 + 157x – 156x – 12246 = 0
 x (2x + 157) – 78 (2x + 157) = 0

(2x + 157) (x – 78) = 0
 2x + 157 = 0
or
(x – 78) = 0
 2x + 157 = 0
or
x – 78 = 0
– 157
 x=
or
x = 78
2
– 157
cannot be acceptable because divisor cannot be negative.
x =
2
 x = 78
 The divisor of 6123 is 78.
WORD PROBLEMS BASED ON AGE
EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :
2.
Sol.
Tinu is younger than Pinky by three years. The product of their ages is
180. Find their ages.
(3 marks)
Tinu’s age be ‘x’ years
Pinky’s age is (x + 3) years
As per the given condition,
x (x + 3) = 180

x2 + 3x – 180 = 0

x2 – 12x + 15x – 180 = 0
S C H O O L S E C TI O N
147
MT
ALGEBRA
 x (x – 12) + 15 (x – 12)

(x – 12) (x + 15)

x – 12 = 0

x = 12
 Age cannot be negative.

x

x
And x + 3 = 12 + 3 = 15
=
=
or
or
0
0
x + 15 = 0
x = – 15

=
– 15
12
EDUCARE LTD.
 Tinu’s age is 12 years and Pinky’s is 15 years.
7.
Sol.
The sum of the ages of father and his son is 42 years. The product of
(3 marks)
their ages is 185, find their ages.
Sum of ages of father and son = 42 years
Let father’s age be x years
 Son’s age = (42 – x) years
 The age of his son is (42 – x) years.
As per the given condition,
x (42 – x) = 185

42x – x2 = 185

0 = x2 – 42x + 185

x2 – 42x + 185 = 0
2

x – 5x – 37x + 185 = 0

x (x – 5) – 37 (x – 5) = 0

(x – 5) (x – 37) = 0

x – 5 = 0 or x – 37 = 0

x = 5 or x = 37
If x = 5, father’s age is less than son’s age.
 x 5
Hence x = 37
And 42 – x = 42 – 37 = 5
 The father’s age is 37 years and son’s age 5 years.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :
30.
(xv)
Sol.
148
Solve the following :
1
times the age of my son. Six years ago,
2
my age was twice the square of the age of my son. What is the present
age of my son ?
(4 marks)
Let the present age of my son be x years
Two years ago,
My son’s age = (x – 2) years
1
9
 My age = 4 (x – 2) years =
(x – 2) years
2
2
Six years ago,
My son’s age = (x – 6) years
My age = 2 (x – 6)2 years
As per the given condition,
9
(x – 2) – 2 (x – 6)2 = 4
2
Multiplying throughout by 2 we get,
= 8
9 (x – 2) – 4 (x – 6)2

9x – 18 – 4 (x2 – 12x + 36)
= 8
Two years ago my age was 4
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.

9x – 18 – 4x2 + 48x – 144

– 4x2 + 57x – 162

0
2

4x – 57x + 162 + 8

4x2 – 57x + 170
2

4x – 40x – 17x + 170

4x (x – 10) – 17 (x – 10)

(x – 10) (4x – 17)
 x – 10 = 0 or
4x – 17 = 0
 x = 10
or
4x = 17
17
 x = 10
or
x=
4
=
=
=
=
=
=
=
=
8
8
4x2 – 57x + 162 + 8
0
0
0
0
0
17
The age of son six years ago becomes negative for x =
4
17
 x 
4
Hence x = 10
 The present age of son is 10 years.
WORD PROBLEMS BASED ON GEOMETRIC FIGURES
EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :
3.
Sol.
The length of the rectangle is greater than its breadth by 2 cm. The area
(3 marks)
of the rectangle is 24 sq.cm, find its length and breadth.
Let the breadth of the rectangle be ‘x’ cm.
 It’s length is (x + 2) cm.
As per the given condition.

Area of rectangle = Length × Breadth

24 = (x + 2) × x

24 = x2 + 2x

0 = x2 + 2x – 24

x2 + 2x – 24 = 0
2

x – 4x + 6x – 24 = 0

x (x – 4) + 6(x – 4) = 0

(x – 4) (x + 6) = 0

x – 4 = 0 or x + 6 = 0

x = 4 or x = -6
 The breadth of rectangle cannot be negative.
 x  -6
Hence x = 4 and x + 2 = 4 + 2 = 6
 The length of rectangle is 6 cm and its breadth is 4 cm.
12.
Sol.
A rectangular playground is 420 sq.m. If its length is increases by 7 m
and breadth is decreased by 5 metres, the area remains the same. Find
the length and breadth of the playground ?
(5 marks)
Let the length of a rectangular playground be ‘x’ m.
 The area of playground is 420 sq.m.
420
 It’s breadth is
m
x
New length = (x + 7) m
S C H O O L S E C TI O N
149
MT
ALGEBRA
EDUCARE LTD.
 420

– 5 m
New breadth = 

x
As per the given condition,
 Area of new rectangle = Length × Breadth
New area












9.
Sol.
=
 420

– 5
(x + 7) 

x
 420

5
(x + 7) × 
 x

2940
420 = 420 – 5x +
– 35
x
Multiplying throughout by x, we get;
0 = – 5x2 + 2940 – 35x
2
5x + 35x – 2940 = 0
x2 + 7x – 588 = 0
[Dividing throughout by 5]
2
x – 21x + 28x – 588 = 0
x (x – 21) + 28 (x – 21) = 0
(x – 21) (x + 28) = 0
x – 21 = 0 or x + 28 = 0
x = 21 or x = -28
The length of playground cannot be negative.
x  -28
Hence x = 21
420
420
=
= 20
And
x
21
The length of a rectangular playground is 21 m and its breadth is 20 m.
420
=
The length of one diagonal of a rhombus is less than the second diagonal
by 4 cm. The area of the rhombus is 30 sq.cm. Find the length of the
diagonals.
(4 marks)
Let the length of other diagonal of a rhombus be ‘x’ cm.
 The length of first diagonal is (x + 4) cm.
1
× Product of length of diagonals
Area of rhombus =
2
1
× x × (x + 4)
Area of rhombus =
2
As per the given condition,
1
x (x + 4) = 30
2

x (x + 4) = 60

x2 + 4x – 60 = 0

x2 + 10x – 6x – 60 = 0

x (x + 10) – 6 (x + 10) = 0

(x + 10) (x – 6) = 0

x + 10 = 0 or x – 6 = 0

x = – 10 or x = 6
 The length of diagonal of the rhombus cannot be negative.
 x  – 10
Hence x = 6
And x + 4 = 6 + 4 = 10
 The length of smaller diagonal of a rhombus is 6 cm and bigger diagonal
is 10 cm.
150
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :
30.
(iii)
Sol.
Solve the following :
One diagonal of a rhombus is greater than other by 4 cm. If the area of
(5 marks)
the rhombus is 96 cm2, find the side of the rhombus.
Let the length of smaller diagonal of a rhombus be x cm
 The length of bigger diagonal is (x + 4) cm
As per the given condition,
1

Area of rhombus =
× Diagonal 1 × Diagonal 2
2
As per the given condition,
1

× x × (x + 4) =
96
2
2

x + 4x = 192

x2 + 4x – 192 = 0

x2 – 12x + 16x – 192 = 0
 x (x – 12) + 16 (x – 12) = 0

(x – 12) (x + 16) = 0

x – 12 = 0 or x + 16 = 0

x = 12 or x = – 16
 The length of diagonal of the rhombus cannot be negative
 x  – 16
A
D
Hence x = 12
 x + 4 = 12 + 4 = 16
Considering ABCD a rhombus
O
 l (AC) = 12 cm
 l (BD) = 16 cm
B
C
O is the intersection point of diagonal AC and BD
1
1
l (AO) =
l (AC) =
× 12 = 6 cm
Diagonal of a rhombus are
2
2
perpendicular bisectors of
1
1
l (BO) =
l (BD) =
× 16 = 8 cm
each other
2
2
AOB is a right angled triangle
In right angled AOB,
[By Pythagoras theorem]
[l (AB)]2 = [l (AO)]2 + [l (BO)]2
 [l (AB)]2 = (6)2 + (8)2
 [l (AB)]2 = 36 + 64
 [l (AB)]2 = 100
Taking square root on both the sides we get,
l (AB) = 10 cm
}
 The side of a rhombus is 10 cm.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :
30.
(ix)
Sol.
Solve the following :
The sum of the areas of two squares is 400sq.m. If the difference between
their perimeters is 16 m, find the sides of two square.
(4 marks)
Difference between the perimeters of two squares is 16 m
 The difference between the sides of the same squares is 4 m
Let the side of smaller square be x cm
 The side of bigger square is (x + 4) cm
S C H O O L S E C TI O N
151
MT
ALGEBRA











EDUCARE LTD.
Area of square = (side) 2
As per the given condition,
x2 + (x + 4)2 = 400
2
2
x + x + 8x + 16 – 400 = 0
2x2 + 8x – 384 = 0
Dividing throughout by 2 we get,
x2 + 4x – 192 = 0
2
x – 12x + 16x – 192 = 0
x (x – 12) + 16 (x – 12) = 0
(x – 12) (x + 16) = 0
x – 12 = 0 or x + 16 = 0
x = 12 or x = – 16
The side of square cannot be negative
x  – 16
Hence x = 12
x + 4 = 12 + 4 = 16
 The side of smaller square is 12 m and bigger square is 16 m.
(xi)
Sol.
Exterior angle of a regular polygon having n-sides is more that of the
polygon having n2 sides by 500 . Find the number of the sides of each
(5 marks)
polygon.
Number of sides of one of the regular polygon = n
Number of sides of the other regular polygon = n2
 360 
 Exterior angle of a polygon having ‘n’ sides = 

 n 
o
 360 
Exterior angle of a polygon having ‘n2’ sides =  2 
 n 
As per the given condition,
360 360
– 2
= 50
n
n
1 
1

360  – 2 
= 50
n n 









n –1
n2
n –1
n2
36 (n – 1)
36n – 36
0
2
5n – 36n + 36
2
5n – 30n – 6n + 36
5n (n – 6) – 6 (n – 6)
(n – 6) (5n – 6)
n–6=0
n=6
=
=
=
=
=
=
=
or
or

n=6
or


152
=
=
o
50
360
5
36
5 (n2)
5n2
5n2 – 36n + 36
0
0
0
0
5n – 6 = 0
5n = 6
6
n=
5
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
 ‘n’ is the number of sides of any polygon
6
 n 
5
Hence n = 6
 n2 = (6)2 = 36
 The number of sides of required regular polygons are 6 and
36 cm respectively.
(xiv)
Sol.
The sides of one regular hexagon is larger than that of the other regular
hexagon by 1cm . If the product of their areas is 243, than find the
sides of both the regular hexangons.
(5 marks)
Let the side of the smaller regular hexagon be x cm.
 The side of the bigger regular hexagon is (x + 1) cm
 Area of regular hexagon =
3 3
× (side)2
2
As per the given condition,














3 3 2
3 3
x +
(x + 1)2
2
2
=
243
93
× x2 (x + 1)2
4
=
243
4
243 × 9  3
[(x2 + x)]2 = 36
Taking square root on both the sides we get,
x (x + 1) = + 6
x (x + 1) = – 6 is not acceptable because product of side of hexagon cannot
be negative
x (x + 1) = 6
x2 + x = 6
x2 + x – 6 = 0
2
x + 3x – 2x – 6 = 0
x (x + 3) – 2 (x + 3) = 0
(x + 3) (x – 2) = 0
x + 3 = 0 or x – 2 = 0
x = – 3 or x = 2
x is a side of regular hexagon
x –3
Hence x = 2
x+1=2+1=3
[x (x + 1)]2
=
 The sides of both the regular hexagon are 2 cm and 3 cm respectively.
(xxii) Around a square pool there is a footpath of width 2km. if the area of the
5
times that of the pool. Find the area of the pool.(5 marks)
footpath is
4
Sol.
Let the side of inner square i.e. pool be x m.
 The width of foot path around the pool is 2 m
 The side of outer square is (x + 2) m

Area of square = side 2
S C H O O L S E C TI O N
153
MT
ALGEBRA












EDUCARE LTD.
As per the given condition,
(Area of outer square) = (Area of inner square) + (Area of footpath)
5 2
2 km
(x + 4)2 = x2 +
x
4
2
2
5
x2 + 8x + 16 = x2 + x2 km
km
4
(x+ 4)km
5 2
8x + 16 =
x
4
Multiplying throughout by 4 we get,
footpath
2 km
32x + 64 = 5x2
(x+ 4)km
5x2 – 32x – 64 = 0
2
5x – 40x + 8x – 64 = 0
5x (x – 8) + 8 (x – 8) = 0
(x – 8) (5x + 8) = 0
x – 8 = 0 or 5x + 8 = 0
x = 8 or 5x = – 8
–8
x = 8 or x =
5
–8
x =
is not acceptable because side of pool cannot be negative.
5
x = 8
x2 = 8 2
x2 = 64
 Area of pool is 64 sq. m.
(ii)
Sol.
In garden there are some rows and columns. The number of trees in a
row is greater than that in each column by 10. Find the number of trees
in each row if the total number of trees are 200.
(3 marks)
Let the number of trees in each column be x
 The number of trees in each column is x + 10
As per the given condition,
x (x + 10) = 200

x2 + 10x – 200 = 0

x2 + 20x – 10x – 200 = 0
 x (x + 20) – 10 (x + 20) = 0

(x + 20) (x – 10) = 0

x + 20 = 0 or x – 10 = 0

x = –20 or x = 10
 The number of trees cannot be negative
 x  – 20
Hence
x = 10
 x + 10 = 10 + 10 = 20
 The number of trees in each row is 20.
WORD PROBLEMS BASED ON SPEED, DISTANCE AND TIME
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 168) :
30.
(xx)
Sol.
154
Solve the following :
From the same place at 7 am ‘A’ started walking in the north at the speed
of 5 km/hr. After 1 hour B started cycling in the east at a speed of
16 km/hr. At what time they will be at distance of 52 km apart from each
(5 marks)
other.
Let O be the point from where A and B started their journey.
S C H O O L S E C TI O N
ALGEBRA
EDUCARE LTD.

















A reached at point N while B reached at point E and distance NE is
52 km
Let the time taken by B to reach at point E be x hrs
The time taken by A to reach at point N (x + 1) hrs.
Distance = Speed × Time
As per the given condition,
In right angled NOE,
= (ON)2 + (OE)2 [By Pythagoras theorem]
(NE) 2
2
(52)
= [5 (x + 1)]2 + (16x)2
2704
= (5x + 5)2 + (16x)2
0
= 25x2 + 50x + 25 + 256x2 – 2704
0
= 281x2 + 50x – 2679 = 0
281x2 – 843x + 893x – 2679 = 0
N
281x (x – 3) + 893 (x – 3)
= 0
(x – 3) (281x + 893)
= 0
x–3=0
or 281x + 893 = 0
52 km
x=3
or 281x = – 893
– 893
x=3
or x =
281
x is the time taken by A
16x km
O
E
– 893
x 
281
Hence x = 3
x+1=4
A took 4 hrs to reach at point N from point O
He started his journey at 7 a.m.
He will be at point N at 11 a.m.
Hence B will be at point E at 11 a.m.
5 (x + 1) km
MT
 A and B will be at distance of 52 km apart from each other at 11 a.m.
(xvii) A car covers a distance of 240km with some speed is increased by 20
km/hr, it will cover the same distance in 2 hours less. find the speed of
the car.
(5 marks)
Sol.
Let the original speed of car be x km/hr.
 Distance covered is 240 km
Dis tance
Speed =
Time
Dis tance
 Time =
Speed
 Time taken by car =
New speed of car =
 240 

 hrs
x 
(x + 20) km/hr
 240 
hrs
 New time taken by car = 
 x  20 
As per the given condition,
240
240
–
= 2
x
x  20
1
1 
240  –
 = 2

 x x  20 
S C H O O L S E C TI O N
155
MT
ALGEBRA













x  20 – x
=
x (x  20)
20
=
2
x  20x
20 (120) =
2400 =
0 =
x2 + 20x – 2400 =
x2 + 60x – 40x – 2400 =
x (x + 60) – 40 (x + 60) =
(x + 60) (x – 40) =
x + 60 = 0 or
x = – 60 or
The speed of car can never
x  – 60
Hence x = 40
EDUCARE LTD.
2
240
1
120
1 (x2 + 20x)
x2 + 20x
x2 + 20x – 2400
0
0
0
0
x – 40 = 0
x = 40
be negative.
 The original speed of car is 40 km/hr.
(v)
Sol.
A man riding on a bicycle cover a distance of 60 km in a direction of
wind and comes back to his original position in 8 hours. If the speed of
the wind is 10 km/hr. Find the speed of the bicycle.
(5 marks)
Let the speed of bicycle be x km / hr
Speed of wind is 10 km /hr
 Speed of bicycle in the direction of wind = (x + 10) km/hr
Speed of the bicycle against the direction of wind = (x – 10) km/hr
Dis tance
Also, Speed =
Time
Dis tance
 Time = Speed
 60 
Time taken by man while riding in the direction of wind =  x  10  hrs


 60 
Time taken by man while riding against the direction of wind =  x – 10  hrs


As per the given condition,
60
60

= 8
x  10 x – 10








156
 1
1 

60 

= 8
 x  10 x – 10 
x – 10  x  10
8
=
(x  10) (x – 10)
60
2x
2
=
2
x – 100
15
Dividing throughout by 2 we get,
x
1
=
2
x – 100
15
x2 – 100 = 15x
x2 – 15x – 100 = 0
2
x – 20x + 5x – 100 = 0
x (x – 20) + 5 (x – 20) = 0
(x – 20) (x + 5) = 0
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.

x – 20 = 0 or x + 5 = 0

x = 20 or x = – 5
 The speed of bicycle cannot be negative
 x –5
Hence x = 20
 The speed of bicycle is 20 km / hr.
(xviii) An express train takes 30 min less for a journey of 440 km, if its usual
speed is increased by 8 km/hr. find its usual speed.
(5 marks)
Sol.
Let the usual speed of train be x km/hr.
 Distance = Time × Speed
Dis tance
Speed =
Time
Dis tance
 Time =
Speed
Distance covered by train = 440 km
 440 
 Time taken by train = 
 hrs
 x 
New speed of train = (x + 8) km/hr
 440 
 New time taken by train =  x  8  hrs.


As per the given condition,
440
440
1 

1
–
 30 min  hr 
=

x
x8
2 
2

1
1 
1
440  –

=

x  8
2
x
x8– x
1
1


=
x (x  8)
2 440
8
1

=
x 2  8x
880

8 (880) = 1 (x2 + 8x)

7040 = x2 + 8x

0 = x2 + 8x – 7040 = 0

x2 + 88x – 80x – 7040 = 0
 x (x + 88) – 80 (x + 88) = 0

(x + 88) (x – 80) = 0

x + 88 = 0 or x – 80 = 0

x = – 88 or x = 80
 The speed of train can never be negative.
 x  – 88
Hence x = 80
 The usual speed of train is 80 km/hr.
WORD PROBLEMS BASED ON COST
EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :
13.
Sol.
The cost of bananas is increased by Re. 1 per dozen, one can get 2 dozen less
for Rs. 840. Find the original cost of one dozen of banana.
(5 marks)
Let the cost of banana per dozen be Rs. x.
Amount for which bananas are bought = Rs. 840
S C H O O L S E C TI O N
157
MT
ALGEBRA
EDUCARE LTD.
840
x
New cost of banana per dozen = Rs. (x + 1)
No. of dozens of bananas for Rs 840 is
840
New No. of dozens of bananas for Rs 840 = x  1
As per the given condition,
840
840
–
2
x
x 1 =












1
1 
840  –
x  1 
x
x  1 – x
840 

 x (x  1) 
=
0
=
2
 1 
840  2
 =
x  x
840 =
2x2 + 2x – 840 =
Dividing throughout by 2,
2x2 + x – 420 =
2
x – 20x + 21x – 420 =
x (x – 20) + 21 (x – 20) =
x – 20 = 0 or
x = 20 or
The cost of bananas cannot
x  –21
Hence x = 20
2
2 (x2 + x)
0
0
0
0
x + 21 = 0
x = –21
be negative.
 The original cost of one dozen banana is Rs. 20.
WROD PROBLEMS BASED ON WORK
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :
30.
(vii)
Sol.
158
Solve the following :
For doing some work Ganesh takes 10 days more than John. If both
work together they complete the work in 12 days. Find the number of
days if Ganesh worked alone?
(5 marks)
Let the number of days required by John alone to complete the
work be x days and Ganesh alone is (x + 10) days
 No. of days requred by ganesh along is (x + 10 ) days.
Also number of days required by both to complete the same work is
12 days
1
 Work done by John in 1 day =
x
1
 Work done by Ganesh in 1 day = x + 10
1
 Work done by both in 1 day =
12
As per the given condition,
1
1
1
+
=
x x + 10
12
x + 10 + x
1

=
x (x + 10)
12
S C H O O L S E C TI O N
MT
EDUCARE LTD.













ALGEBRA
2x + 10
1
=
2
x + 10x
12
12 (2x + 10) = 1 (x2 + 10x)
24x + 120 = x2 + 10x
0 = x2 + 10x – 24x – 120
2
x – 14x – 120 = 0
x2 – 20x + 6x – 120 = 0
x (x – 20) + 6 (x – 20) = 0
(x – 20) (x + 6) = 0
x – 20 = 0 or x + 6 = 0
x = 20 or x = – 6
The numbers of days cannot be negative
x –6
Hence x = 20
x + 10 = 20 + 10 = 30
 Ganesh alone worked for 30 days.
(xii)
Sol.
Tinu takes 9 days more than his father to do a certain piece of work.
Together they can do the work in 6 days. How many days will tinu take
to do that work.
(5 marks)
Let the number of days required by father alone to do a certain
piece of work be ‘x’ days and by Tinu alone is (x + 9) days
 No. of days required by tinu along is (x + 9) days
Also number of days required by both to complete the same work is 6 days.
1
Work done by father in 1 day =
x
1
Work done by Tinu in 1 day = x + 9
1
Work done by both in 1 day =
6
As per the given condition,
1
1
1
+
=
x
x+ 9
6
x+ 9+ x
1

=
x (x + 9)
6
2x + 9
1

=
x 2 + 9x
6

6 (2x + 9) = 1 (x2 + 9x)

12x + 54 = x2 + 9x

0 = x2 + 9x – 12x – 54

0 = x2 – 3x – 54
2

x – 3x – 54 = 0

x2 – 9x + 6x – 54 = 0

x (x – 9) + 6 (x – 9) = 0

(x – 9) (x + 6) = 0

x – 9 = 0 or x + 6 = 0

x = 9 or x = – 6
 The number of days cannot be negative
 x 6
Hence x = 9
 x + 9 = 9 + 9 = 18
 Tinu alone requires 18 days to complete the work.
S C H O O L S E C TI O N
159
MT
ALGEBRA
(xxi)
Sol.
EDUCARE LTD.
One tank can be filled up by two taps in 6 hours. The smaller tap alone
takes 5 hours more than the bigger tap alone. Find the time required by
(5 marks)
each tap to fill the tank separately.
Let the time taken to fill a tank by a bigger tap alone be x hrs.
 The time taken by smaller tap alone is (x + 5) hrs.
Time taken by both the taps together to fill the same tank is 6 hrs.
1
 Portion of tank filled in 1 hr by bigger tap =
x
1
Portion of tank filled in 1 hr by smaller tap = x  5
1
Portion of tank filled in 1 hr by both taps together =
6
As per the given condition,
1
1
1
+ x5
=
x
6
x5x
1

=
x (x  5)
6
2x  5
1

=
x 2  5x
6

6 (2x + 5) = 1 (x2 + 5x)

12x + 30 = x2 + 5x

0 = x2 + 5x – 12x – 30
2

x + 3x – 10x – 30 = 0

x (x + 3) – 10 (x + 3) = 0

(x + 3) (x – 10) = 0

x + 3 = 0 or x – 10 = 0

x = – 3 or x = 10
 x is the time taken bybigger tap
 x  – 3
Hence x = 10
 x + 5 = 10 + 5 = 15
 Time taken by bigger tap alone is 10 hrs and smaller tap alone is 15 hrs.
MCQ’s
1.
2.
The sum of roots ( + ) = ............. .
b
(b)
(a) –
a
c
(c) –
(d)
a
In formula method the value of x = ............. .
(a)
(c)
3.
160
b
a
c
a
b –
b2 – 4ac
2a
(b)
b 
b2 – 4ac
2a
(d)
–b 
b2 – 4ac
2a
b2  4ac
2a
In complemiting square method the formula of third term is ............. .
1
(a) 2 × (coefficient of x2)
(b)
× (coefficient of x2)
2
2
1

(d) (2 × coefficient of x)2
(c)   coefficient of x 
2

S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
4.
The product of the roots ( . ) = ............. .
b
c
(b) –
(a) –
a
a
b
c
(c)
(d)
a
a
5.
The solution set of y2 – 16y + 63 = 0 are = ............. .
(a) – 9 and – 7
(b) – 9 and 7
(c) 9 and – 7
(d) 9 and 7
6.
The factors of y2 – 5y – 24 = 0 is = ............. .
(a) – 8 and – 3
(b) – 8 and 3
(c) 8 and – 3
(d) 8 and 3
7.
The roots of the quadratic equation x2 + 5x – 14 is = ............. .
(b) – 7
(a)
2
(c) – 3
(d) – 2
8.
Which of these following is quadratic equation ............. .
5
 4m  5
(a) m –
(b) n3 – n + 4 = n3
m
5
– 3  x2
(c)
(d) 13 = – 5y2 – y3
x
9.
If the roots of quadratic equation are real and equal then  must be ............. .
(a) equal to zero
(b) greater then zero
(c) greater than one
(d) equal to one
10.
If m2 – 36 = 0 then m2 is equal to ............. .
(a) 6
(b) 36
(c) 9
(d) – 6
11.
Three times the square of natural number is 363 is written in the
mathematical equation form as ............. .
(a) x2 + 3 = 363
(b) x2 – 3 = 363
x2
 363
(c) 3x2 = 363
(d)
3
12.
Which of these following is not an quadratic equation ............. .
(a) –
(c)
5 2
x  2x  9
3
5
– 3  x2
x
(b)
(x + 3) (x + 4) = 0
(d)
m–
5
= 4m + 5
m
13.
If one root of quadratic equation is kx2 – 7x + 12 = 0 is 3 then value of k is
............. .
(a) – 1
(b) 1
(c) 3
(d) none of these
14.
If one root of quadratic equation is 1 – 3 then product of root is ............. .
(a) 2
(b) – 2
(c) – 44
(d) 44
S C H O O L S E C TI O N
161
MT
ALGEBRA
EDUCARE LTD.
15.
The quadratic equation is in form of ............. .
(a) x2 + (sum of roots)x – (product of roots) = 0
(b) x2 – (sum of roots)x + (product of roots) = 0
(c) x2 + (product of roots)x – (sum of roots) = 0
(d) x2 – (product of roots) x + (sum of roots) = 0
16.
If one root of quadratic equation is 4 2 – 3 then sum of roots is ............. .
(a) – 8 2
(b) 8 2
(c) 23
(d) – 6
17.
If sum of natural number and its square is 30, then quadratic equation is
written as ............. .
(a) x2 – x – 30 = 0
(b) x2 + x – 30 = 0
(c) x2 – x + 30 = 0
(d) x2 + x + 30 = 0
18.
x2 +
1
= ............. .
x2
2
2
1 

(a)  x – 2  – 2
x 

(b)
1 

x  2   2
x 

(d)
none of these
3 + 3 = ............. .
(a) ( + )3 – 3 ( + )
(c) ( + )3 – 3 ( – )
(b)
(d)
( – )3 + 3 ( – )
( – )3 – 3 ( – )
( – )2 = ............. .
(a) ( + )2 + 4
(c) ( + )2 – 4
(b)
(d)
( + )2 – 3
( + )2 + 2
2
1

(c)  x –   2
x

19.
20.
: ANSWERS :
1.
(a) –
b
a
2.
5.
1

(c)   coefficient of x 
2


(d) 9 and 7
7.
(b)
9.
(a)
11.
(c)
13.
15.
16.
(b)
(b)
(d)
3.
(b)
–b 
b2 – 4ac
2a
2
c
a
6. (b) – 8 and 3
5
 4m  5
–7
8. (a) m –
m
equal to zero
10. (b) 36
5
– 3  x2
3x2 = 363
12. (c)
x
1
14. (b) – 2
2
x – (sum of roots)x + (product of roots) = 0
–6
17. (b) x2 + x – 30 = 0
4.
(d)
2
18.
20.
162
1

(c)  x –   2
x

(c) ( + )2 – 4
19.
(a) ( + )3 – 3 ( + )
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
HOTS PROBLEMS
(Problems for developing Higher Order Thinking Skill)
10.
If the roots of the quadratic equation ax2 + cx + c = 0 are in the ratio
p
q
c
+
+
=0
q
p
a
As we know, as per the given equation
–c
 + 
=
a
c

=
a

p
Also,
=

q
(3 marks)
p : q show that
Sol.
Now,
p

q
q

p
c
a
=
=
=
=
=
=
=

11.
Ans.











+



. 
. 
. 
 +  + 

–c
c

a
a

0

0
p
q
c


0
q
p
a
Hence proved.
If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to
the sum of the squares of their reciprocals then prove that 2a2c = c2b +
b 2a.
(4 marks)
2
Let the root of the quadratic equation ax + bx + c = 0 be  and 
and as we know for the ab are equation
–b
 + 
=
a
c

=
a
As per the given condition,
1
1
 2
 + 
=
2


2  2
 + 
=
2 2
 2  2  2   2 – 2
2
 + 
=
  
S C H O O L S E C TI O N
163
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ALGEBRA
EDUCARE LTD.
     – 2
  2
2
 + 
=
2




12.
Ans.
 b
c
–  – 2  
 a
a
–b
2
=
c
a
 
a
2
2
b c
b
2c
–
–
=
a
a a2
a2
2
2
b
2ac
bc
– 3
– 2
=
2
a
a
a
bc2
b2 – 2ac
– 3
=
a
a2
Multiplying both sides by a3, we get
 bc 2 
 b2 – 2ac 
a3  – 3  = a3  –

a2
 a 


– bc2
= a (b2 – 2ac)
– bc2
= ab2 – 2a2c
2
 2a c
= c 2b + b 2a
Hence proved.
Form the quadratic equation whose roots are the squares of the sum and square
of the difference of the equation 2x2 + 2(m + n)x + m2 + n2 = 0 (4 marks)
Let  and  be the roots of the equation
2x2 + 2 (m + n)x + m2 + n2 = 0 as we know
– 2 (m  n)
m 2  n2
 + 
=
and  =
2
2
Let roots of new quadratic equation be 1 and 1
1 = ( + )2
=
(m + n)2 and  .  =
( – )2
=
( + )2 – 4
m2  n 2
2
 m 2  n2 
m2 + 2mn + n2 – 4 

2

2
2
2
2
= m + 2mn + n – 2m – 2n
= – m2 + 2mn – n2
= – (m2 – 2mn + n2)
1
= (a – b)2 = – (m – n)2
Now, 1 + 1 = m2 + 2mn + n2 – m2 + 2mn – n2
1 + 1= 4mn
1 . 1 = (m + n)2 × – (m – n)2 = – (m2 – n2)2
=
 Quadratic equation is x2 – 4mnx – (m2 – n2)2 = 0.
13.
Sol.
164
Find the condition that the equations ax2 + bx + c = 0 and a1x2 + b1x + c1 = 0
may have a common root. Find this common root.
(5 marks)
ax2 + bx + c = 0
......(i)
a1x2 + b1x + c1 = 0
.....(ii)
Multiplying (i) by a1,
a1ax2 + a1bx + a1c = 0
......(iii)
Multiplying (ii) by a,
a1ax2 + ab1x + ac1 = 0
......(iv)
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
Subtracting (iv) from (iii),
a1ax2 + a1bx + a1c = 0
a1ax2 + ab1x + ac1 = 0
(–)
(–)
(–)
a1bx – ab1x + a1c – ac1 = 0
 x (a1b – ab1) = ac1 – a1c
ac1 – a1c
 x = a b – ab
1
1
ac1 – a1c
Substituting x = a b – ab in (i),
1
1
2
 ac – a1c 
 ac1 – a1c 
a 1
 b
c =0
 a1b – ab1 
 a1b – ab1 
2
a ac1 – a1c 
b ac1 – a1c 

c =0

2
a1b – ab1
a1b – ab1 
Multiplying throughout by (a1b – ab1)2 we get,
a (ac1 – a1c)2 + b (ac1 – a1c) (a1b – ab1)2 + c(a1b – ab1)2 = 0
The condition for the equations to have common root is
 a (ac1 – a1c)2 + b (ac1 – a1c) (a1b – ab1)2 + c (a1b – ab1)2 = 0
14.
Sol.
Find the value if p of the
have a common root.
6x2 – 17x + 12 = 0

6x2 – 9x – 8x + 12
 3x (2x – 3) – 4 (2x – 3)

(2x – 3) (3x – 4)

2x – 3 = 0 or 3x – 4

equations 3x2 – 2x + p and 6x2 – 17x + 12 = 0
(4 marks)
=
=
=
=
0
0
0
0
3
4
or x =
2
3
The first equation 3x2 – 2x + p = 0
If x = 3 is a common root then this value of ‘x’ will satisfy the equation
x=
2




3
3
3   –2 
2
 
2
9
3 –3
4
9
–3
2
3
2

15.
Sol.
p
=
0
p
=
0
p
=
0
p
=
0
p
=
–
3
2
If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal
to the sum of the squares of their reciprocals. Show that bc2, ca2, ab2
are in A.P.
(5 marks)
To show bc2, ca2, ab2 are in A.P.
i.e. to prove ca2 – bc2 = ab2 – ca2
i.e. ca2 + ca2 = ab2 + bc2
i.e. 2ca2 = ab2 + bc2
.......(i)
Let the roots quadratic equation ax 2 + bx + x = 0 be a and b as we
know for the above equation
S C H O O L S E C TI O N
165
MT
ALGEBRA
EDUCARE LTD.
–b
a
c

=
a
As per the given condition,
1
1

 + 
=
2 2
 + 
=

 + 
=

 + 
=
2  2
2 2
 2  2 2   2 – 2

 + 
  
2
     – 2
  2
2
=
2

–b
a
=

– b c2
a a2
=







34.
Sol.
166
– bc 2
=
a3
– bc 2
=
a3
Multiplying both sides
 – bc 2 
a3  3 
 a 
– bc2
– bc2
– bc2
a2c – bc2
2
ca – bc2
=
=
=
=
=
=
 –b
c

 –2 
a


a
2
c
 
a
2
b
2c
–
2
a
c
b2
2ac
– 2
2
a
a
2
b – 2ac
a2
by a3 we get,
 b2 – 2ac 
a3 

a2


a (b2 – 2ac)
ab2 – 2a2c
ab2 – a2c – a2c
ab2 – a2c
ab2 – ca2
bc2, ca2, ab2 are in A.P.
Solve : 216x6 – 793x3 + 216 = 0
216x6 – 793x3 + 216 = 0
216 (x3)2 – 793x3 + 216 = 0
Substituting x3 = a we get,
216a2 – 793a + 216 = 0
 216a2 – 64a – 729a + 216 = 0
 216a2 – 64a – 729a + 216 = 0
 8a (27a – 8) – 27 (27a – 8) = 0

(27a – 8) (8a – 27) = 0
 27a – 8 = 0
or

27a = 8
or
8

a =
or
27
(5 marks)
8a – 27 = 0
8a = 27
27
a =
8
S C H O O L S E C TI O N
MT
ALGEBRA
EDUCARE LTD.
Resubstituting a = x3
8
x3 =
or
27
Taking cube roots throughout,
2
or
x =
3
x3
=
27
8
x
=
3
2
2 3 
 Solution set =  , 
3 2 
35.
Sol.
A man travels by boat 36 km down a river and back in 8 hours. If the
speed of his boat in still water is 12 km per hour, find the speed of the
river curved.
(4 marks)
Speed of boat in still water
=
12 km/hr.
Let speed of river current
=
x km/hr.
 Speed of boat up the river
=
(12 – x) km/hr.
Speed of boat down the river =
(12 + x) km/hr.
Dis tance
Time = Speed
 36 
 Time taken by boat to travel 36 km down the river =  12  x  hrs.


As per the given condition,
36
36
+
12 + x 12 – x







12 – x  12  x 
36 

 (12  x) (12 – x) 


24
36 
2 
144 – x 
36 × 24
8
108
x2
x2
x
 1
1 
= 8 36 12  x  12 – x  = 8


= 8
= 8
= 144 – x2
=
=
=
=
144 – x2
144 – 108
36
+6
[Taking square roots]
 x = – 6 is not acceptable because speed cannot be negative.
38.
Sol.
A business man bought some items for Rs. 600, keeping 10 items for
himself he sold the remaining items at a profit of Rs. 5 per item. From
the amount received in this deal he could buy 15 more items. Find the
(5 marks)
original price of each item.
Let the original price of each item be Rs. x
Total cost price = Rs. 600
600
 No. of items bought =
x
600
– 10
No. of items sold after keeping 10 items =
x
Selling price each item = Rs. (x + 5)
 600

– 10 
 Total selling price = (x + 5) 
 x

S C H O O L S E C TI O N
167
MT
ALGEBRA
EDUCARE LTD.
 600

– 10  – 600
Net profit made (x + 5) 
 x

But profit is equal to cost of 15 items = Rs. x
As per the given condition,
 600

– 10  – 600
(x + 5) 
 x












= 15x
 600

 600

x
– 10  + 
– 10  – 600 = 15x
 x

 x

3000
600 – 10x +
– 50 – 600 = 15x
x
3000
– 10x +
– 50 – 15x = 0
x
3000
– 25x +
– 50 = 0
x
Multiplying throughout by x,
– 25x + 3000 – 50x = 0
25x2 – 50x + 3000 = 0
Dividing throughout by – 25 we get,
x2 + 2x – 120 = 0
x2 + 12x – 10x – 120 = 0
x (x + 12) – 10 (x + 12) = 0
(x + 12) (x – 10) = 0
x + 12 = 0
or
x – 10 = 0
x = – 12
or
x = 10
x = – 12 is not acceptable because cost cannot be negative.
x = 10
 Original price of each items is Rs. 10.
  
168
S C H O O L S E C TI O N
S.S.C.
Marks : 30
CHAPTER 2 : Quadratic Equations
ALGEBRA
SET - A
Duration : 1 hr.
Q.1. Atttempt any TWO of the following :
2
(i)
State whether the following equation is a (y – 2) (y + 2) = 0.
(ii)
Write the following quadratic equations in standard form : (m + 4)
(m – 10) = 0.
(iii)
Find the value of discriminant of of the following equation :
3x2 + 2x – 1 = 0.
Q.2. Attempt any TWO of the following :
4
(i)
If one root of the quadratic equation x2 – 7x + k = 0 is 4, then find
the value of k.
(ii)
Solve the following quadratic equation by factorization method :
x2 – 13x – 30 = 0.
(iii)
Solve the following quardratic equation by factorization method :
y2 – 3 = 0.
Q.3. Attempt any TWO of the following :
6
(i)
Solve the following quadratic equation by completing square :
x2 + 8x + 9 = 0.
(ii)
Form the quadratic equation if its roots are – 2 and
(iii)
Find the value of k for which given equation has real and equal roots :
(k – 12)x2 + 2 (k – 12)x + 2 = 0
11
.
2
Q.4. Attempt any TWO of the following :
(i)
Find m, if the roots of the quadratic equation (m – 1) x 2 – 2
(m – 1) x + 1 = 0 has real and equal roots.
(ii)
If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the
other, find k.
(iii)
Solve the following equation : x2 +
12
= 7.
x2
Q.5. Attempt any TWO of the following :
(i)
8
1
 2 1 

Solve the following equation : 2  x + 2  – 9  x +  + 14 = 0
x 
x


(ii)
The product of four consecutive positive integers is 840. find the
largest number.
(iii)
If the difference of the roots of the quadratic equation is 3 and difference
between their cubes is 189, find the quadratic equation.
Best Of Luck 
10
S.S.C.
Marks : 30
CHAPTER 2 : Quadratic Equations
ALGEBRA
SET - A
Duration : 1 hr.
Q.1. Atttempt any TWO of the following :
2
(i)
State whether following is a quadratic equation 3y2 – 7 =
3 y.
(ii)
Write the following quadratic equation in standard form (m + 4)
(m – 10) = 0.
(iii)
Find the value of discriminant of each of the following equation :
3x2 + 2x – 1 = 0.
Q.2. Attempt any TWO of the following :
4
(i)
Solve the following quadratic equation by factorization method :
x2 + 10x + 24 = 0.
(ii)
If one root of the quadratic equation 3y2 – ky + 8 = 0 is
the value of k.
(iii)
2
, then find
3
Solve the following quadratic equation by factorization method :
49x2 = 36.
Q.3. Attempt any TWO of the following :
6
(i)
Solve the following quadratic equation using formula :
x2 + 3x – 10 = 0.
(ii)
Solve the following quadratic equation using formula :
2x2 + 5x – 2 = 0.
(iii)
From the quadratic equation whose roots are
3
–2
and
.
4
3
Q.4. Attempt any TWO of the following :
(i)
Find the value of k for which given equation has real and equal
roots : (k – 12)x2 + 2 (k – 12)x + 2 = 0
(ii)
Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are in
the ratio 2 : 5.
(iii)
12
Solve : 35y2 + y 2 = 44.
Q.5. Attempt any TWO of the following :
(i)
8
Solve the following equation :
1 
1


12  x 2 + 2  – 56  x +  + 89 = 0
x
x 


(ii)
A rectangular playground is 420 sq.m. If its length is increases by
7 m and breadth is decreased by 5 metres, the area remains the
same. Find the length and breadth of the playground ?
(iii)
If the difference of the roots of the quadratic equation is 5 and the
difference of their cubes is 215, find the quadratic equation.
Best Of Luck 
10