Quadratic equations
1. Introduction
Equations of the form ax + b = 0 where a 6= 0 are called linear equations and have only one
solution.
2
For example, 3x − 2 = 0 is the linear equation with a = 3 and b = −2. It has the solution x = .
3
Equations of the form ax2 + bx + c = 0 where a 6= 0 are called quadratic equations. They
may have two, one or zero solutions.
Here are some simple quadratic equations which clearly show the truth of this statement:
Equation
x2 − 4 = 0
(x − 2)2 = 0
x2 + 4 = 0
ax2 + bx + c = 0 form
x2 + 0x − 4 = 0
x2 + 4x + 4 = 0
x2 + 0x + 4 = 0
a
1
1
1
b
0
−4
0
c
−4
4
4
Solutions
x = 2 or x = −2
x=2
none as x2 is always ≥ 0
two
one
zero
Now consider the example x2 + 3x − 10 = 0.
If x = 2, then x2 + 3x − 10 = 22 + 3 · 2 − 10 = 0, and
if x = −5, then x2 + 3x − 10 = (−5)2 + 3 · (−5) − 10 = 0
x = 2 and x = −5 both satisfy the equation x2 + 3x − 10 = 0, so we say that they are both
solutions.
In this lesson we will learn methods for solving quadratic equations without using trial and error, and
apply them to practical problems.
2. Quadratic equations of the form x2 = k
Consider the equation x2 = 7.
√ √
√
Now 7 · 7 = 7, so x = 7 is one solution,
√
√
√
and (− 7) · (− 7) = 7, so x = − 7 is also a solution.
√
√
Thus, if x2 = 7, then x = ± 7.
(± 7 is read as “plus or minus the square root of 7”).
Solution of x2 = k

√
 x=± k
If x2 = k then
x=0

there are no real solutions
if k > 0
if k = 0
if k < 0
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2
Example 1
Solve for x:
a) 2x2 + 1 = 15
b) 2 − 3x2 = 8
2x2 + 1 = 15
a)
∴ 2x2 = 14
∴ x2 = 7
√
∴ x= ± 7
b)
∴
(take 1 from both sides)
(divide both sides by 2)
2 − 3x2 = 8
−3x2 = 6
∴ x2 = −2
(take 2 from both sides)
(divide both sides by − 3)
which has no solutions as x2 cannot be < 0.
Example 2
Solve for x:
a)
a) (x − 3)2 = 16
(x − 3)2 = 16
√
∴ x − 3 = ± 16
∴
x − 3 = ±4
∴
∴
x = 3±4
x = 7 or − 1
b) (x + 2)2 = 11
b)
(x + 2)2 = 11
√
∴ x + 2 = ± 11
√
∴ x = −2 ± 11
For equations of the form (x ± a)2 = k we do not expand the LHS.
Exercises - Set A
1. Solve for x:
a) x2 = 100
b) 5x2 = 20
c) 6x2 = 54
e) 7x2 = 0
f) 3x2 − 2 = 25
g) 4 − 2x2 = 12
d) 5x2 = −45
h) 4x2 + 2 = 10
2. Solve for x:
a) (x − 1)2 = 9
e) (x + 2)2 = 0
b) (x + 4)2 = 16
f) (2x − 5)2 = 0
c) (x + 2)2 = −1
g) (3x + 2)2 = 4
d) (x − 4)2 = 5
h) 13 (2x + 3)2 = 2
3. The Null Factor law
For quadratic equations which are not of the form x2 = k, we need other methods of solution. One
method is to factorise the quadratic and then apply the Null factor law.
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The Null Factor law states that:
When the product of two (or more) numbers is zero, then at least one of them must be zero.
So, if ab = 0 then a = 0 or b = 0.
This is specially useful to solve quadratic equations of the form ax2 + bx = 0.
Example 3
a) 3x2 − 15x = 0
Solve for x:
b)
3x2 − 15x = 0
3x(x − 5) = 0
a)
∴
∴
b) (x − 4)(3x + 7) = 0
∴
3x = 0 or x − 5 = 0
∴ x = 0 or x = 5
(x − 4)(3x + 7) = 0
x − 4 = 0 or 3x + 7 = 0
∴
x = 4 or 3x = −7
∴
x=4
or x = − 73
Exercises - Set B
1. Solve for the unknown using the Null Factor law:
a) 3x = 0
e) abc = 0
b) −7y = 0
2
f) a = 0
c) ab = 0
d) 2xy = 0
g) pqrs = 0
h) a2 b = 0
2. Solve for x using the Null Factor law:
a) x(x − 5) = 0
b) 2x(x + 3) = 0
g) (2x + 1)(2x − 1) = 0
h) 11(x + 2)(x − 7) = 0
d) 3x(7 − x) = 0
2
j) x = 0
e) −2x(x + 1) = 0
2
k) 4(5 − x) = 0
c) (x + 1)(x − 3) = 0
f) 4(x + 6)(2x − 3) = 0
i) −6(x − 5)(3x + 2) = 0
l) −3(3x − 1)2 = 0
Remember that in order to use the Null Factor law we must have one side of the equation equal to
zero.
Example 4
Solve for x:
x2 = 3x
x2 = 3x
∴ x2 − 3x = 0
∴ x(x − 3) = 0
∴
∴
x = 0 or x − 3 = 0
x = 0 or x = 3
rearranging so RHS = 0
factorising the LHS
Null factor law
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Illegal cancelling
Let us reconsider the equation x2 = 3x from the previous example. We notice that there is a common
factor of x on both sides.
x2
3x
If we cancel x from both sides, we will have
=
and thus x = 3.
x
x
Consequently, we will lose the solution x = 0.
From this example we conclude that:
We must never cancel a variable that is a common factor from both sides of an equation unless
we know that the factor cannot be zero.
Exercises - Set C
1. Solve for x:
a) x2 − 7x = 0
e) 3x2 + 6x = 0
b) x2 − 5x = 0
f) 2x2 − 5x = 0
c) x2 = 8x
d) x2 = 4x
g) 4x2 = 5x
h) 3x2 = 9x
4. Completing the square
A quadratic equation of the type ax2 + bx + c = 0 may not be so easy to factorise, then to solve
this type of equations we need a different technique.
Let us consider for example the equation x2 + 4x − 7 = 0.
√
√
We saw in Example 2 that if (x + 2)2 = 11, then x + 2 = ± 11 and therefore x = −2 ± 11.
So, we can solve x2 +4x−7 = 0 if we can rearrange it so there is a perfect square on the left hand side:
x2 + 4x − 7 = 0
∴ x2 + 4x = 7
x2 + 4x + 4 = 7 + 4
(adding 4 to both sides to complete a perfect square on the LHS)
2
∴ (x + 2) = 11
√
∴ x + 2 = ± 11
√
∴ x = −2 ± 11
√
Hence the solutions to x2 + 4x − 7 = 0 are x = −2 ± 11.
∴
The process of creating a perfect square on the left hand side is called completing the square.
From our previous study of perfect squares we observe that:
(x + 3)2 = x2 + 2 · 3 · x + 32
(x − 5)2 = x2 − 2 · 5 · x + 52
(x + p)2 = x2 + 2 · p · x + p2
So, the constant term is “the square of half the coefficient of x”.
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Example 5
To create a perfect square on the LHS, what must be added to both sides of the equation:
a) x2 + 8x = −5
b) x2 − 6x = 13?
What does the equation become in each case?
8
2
= 4. So, we add 42 to both sides and the
−6
2
= 3. So, we add (−3)2 = 32 to both sides
a) In x2 + 8x = −5, half the coefficient of x is
equation becomes
x2 + 8x + 42 = −5 + 42
∴ (x + 4)2 = −5 + 16
∴
(x + 4)2 = 11
b) In x2 − 6x = 13, half the coefficient of x is
and the equation becomes
x2 − 6x + 32 = 13 + 32
∴
∴
(x − 3)2 = 13 + 9
(x − 3)2 = 22
Example 6
Solve for xby completing the square, leaving answers in simplest radical form:
a) x2 + 2x − 2 = 0
b) x2 − 4x + 6 = 0
a)
∴
x2 + 2x − 2 = 0
∴ x2 + 2x = 2
x2 + 2x + 12 = 2 + 12
move constant term to RHS
add 12 to both sides
(x + 1)2 = 3
factorise LHS, simplify RHS
∴
∴
√
x+1 = ± 3
∴
b)
∴
x = −1 ±
√
3
2
x − 4x + 6 = 0
∴ x2 − 4x = −6
x2 − 4x + 22 = −6 + 22
move constant term to RHS
add 22 to both sides
∴ (x − 2)2 = −2
factorise LHS, simplify RHS
which is impossible as no perfect square can be negative. Hence, no real solutions exist.
Exercises - Set D
1. For each of the following equations find what must be added to both sides of the equation
to create a perfect square on the LHS, and write each equation in the form (x + p)2 = k:
a) x2 + 2x = 5
d) x2 − 6x = −3
g) x2 + 12x = 13
b) x2 − 2x = −7
c) x2 + 6x = 2
h) x2 + 5x = −2
i) x2 − 7x = 4
e) x2 + 10x = 1
f) x2 − 8x = 5
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Exercises - Set E
1. If possible, solve for x by completing the square, leaving answers in simplest radical form:
a) x2 − 4x + 1 = 0
b) x2 − 2x − 2 = 0
c) x2 − 4x − 3 = 0
g) x2 + 6x + 3 = 0
h) x2 − 6x + 11 = 0
i) x2 + 8x + 14 = 0
n) x2 + 3x − 1 = 0
o) x2 + 5x − 2 = 0
d) x2 + 2x − 1 = 0
j) x2 + 3x + 2 = 0
m) x2 + x − 1 = 0
e) x2 + 2x + 4 = 0
k) x2 = 4x + 8
f) x2 + 4x + 1 = 0
l) x2 − 5x + 6 = 0
5. The quadratic formula
Many quadratic equations cannot be solved by factorisation, and completing the square is rather
tedious. Consequently, the quadratic formula has been developed.
√
−b ± b2 − 4ac
2
If ax + bx + c = 0 where a 6= 0, then x =
.
2a
If ax2 + bx + c = 0
Proof:
then
b
c
x2 + x + = 0
a
a
b
x2 + x
a
2
b
b
x2 + x +
a
2a
2
b
∴
x+
2a
2
b
∴
x+
2a
∴
∴
∴
= −
dividing each term by a, as a 6= 0
c
a
c
= − +
a
= −
b
2a
2
completing the square on LHS
c · 4a
b2
+ 2
a · 4a 4a
b2 − 4ac
4a2
r
b
b2 − 4ac
x+
= ±
2a
4a2
√
b
b2 − 4ac
∴ x = − ±
2a
2a
√
−b ± b2 − 4ac
∴ x =
2a
=
Example 7
Solve for x:
x2 − 2x − 2 = 0
x2 − 2x − 2 = 0 has a = 1, b = 2, c = −2, then
p
√
√
√
−(−2) ± (−2)2 − 4(1)(−2)
2± 4+8
2 ± 12
2±2 3
x=
=
=
=
= 1±3
2(1)
2
2
2
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Example 8
Solve for x:
2x2 + 3x − 4 = 0
2x2 + 3x − 4 = 0 has a = 2, b = 3, c = −4, then
p
√
√
−3 ± 32 − 4(2)(−4)
−3 ± 9 + 32
−3 ± 41
=
=
x=
2(2)
4
4
Exercises - Set F
1. Use the quadratic formula to solve for x:
a) x2 + 9x + 14 = 0
b) x2 + 5x + 6 = 0
c) x2 − 3x + 2 = 0
d) x2 − 5x + 6 = 0
e) x2 + 2x = 2
g) x2 = 4x + 1
h) x2 + 1 = 3x
i) x2 + 8x + 5 = 0
j) 2x2 + 7x + 5 = 0
k) 3x2 + 13x + 4 = 0
l) 5x2 = 13x + 6
m) 9x + 36 = x2
n) 2x2 = 2x + 1
o) 9x2 = 6x + 1
p) 25x2 + 1 = 20x
q) 2x2 + 6x + 1 = 0
r) 3x2 + 2x − 2 = 0
f) x2 + 2 = 6x
2. Solve for x by first expanding brackets:
a) x(x + 5) + 2(x + 6) = 0
b) x(1 + x) + x = 3
d) 3x(x + 2) − 5(x − 3) = 17
e) 4x(x + 1) = −1
g) (x + 2)(x − 1) = 5
2
2
h) (x + 1) = 3 − x
c) (x − 1)(x + 9) = 8x
f) 2x(x − 6) = x − 20
3. Solve for x by first eliminating the algebraic fractions:
a)
x
2
=
3
x
b)
x−1
3
=
4
x
c)
x−1
x + 11
=
x
5
d)
2x
1
=
3x + 1
x+2
e)
2x + 1
= 3x
x
f)
x+2
x
=
x−1
2
6. The discriminant, ∆
x2 + 2x + 5 = 0. Using the quadratic formula, the solutions are:
p
√
−2 ± 4 − 4(1)(5)
−2 ± −16
x=
=
2(1)
2
√
However, in the real number system, −16 does not exist. We therefore say that
has no real solutions.
Consider
x2 + 2x + 5 = 0
Consider now x2 + 4x + 4 = 0. Using the quadratic formula, the solutions are:
p
√
−4 ± 16 − 4(1)(4)
−4 ± 0
−4
x=
=
=
= −2
2(1)
2
2
√
As we get 0 = 0, there is only one solution.
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In the quadratic formula, the quantity b2 − 4ac under the square root sign is called the discriminant.
The symbol delta or ∆ is used to represent the discriminant, so ∆ = b2 − 4ac.
Notice that:
• if ∆ = 0, x =
• if ∆ > 0,
• if ∆ < 0,
−b
is the only solution and is known as a repeated root.
2a
√
∆ is a real number and so there are two distinct real roots
√
√
−b + ∆
−b − ∆
and
2a
2a
√
∆ does not exist and so we have no real solution.
Example 9
Use the discriminant to determine the nature of the roots of:
a) 2x2 − 3x + 4 = 0
b) 4x2 − 4x − 1 = 0
a)
b)
∆ = b2 − 4ac = (−3)2 − 4 · 2 · 4 = −23 < 0, so there are no real roots.
∆ = b2 − 4ac = (−4)2 − 4 · 4 · (−1) = 32 > 0, so there are two distinct real roots.
Exercises - Set G
1. Find the discriminant of:
a) x2 − 2x − 7 = 0
b) 2x2 − 3x + 6 = 0
d) 2x2 − 6x − 4 = 0
e) 3x2 + 7x − 1 = 0
c) x2 − 11 = 0
f) 4x2 − 7x + 11 = 0
2. Using the discriminant only, state the nature of the solutions of:
√
a) x2 + 7x − 2 = 0
b) x2 + 4 2x + 8 = 0
c) 2x2 + 3x − 1 = 0
d) 6x2 + 5x − 4 = 0
e) x2 + x + 6 = 0
f) 9x2 + 6x + 1 = 0
3. Show that the following quadratic equations have no real solutions:
a) x2 − 3x + 12 = 0
b) x2 + 2x + 4 = 0
c) −2x2 + x − 1 = 0
4. Solve for x, where possible:
a) x2 − 25 = 0
b) x2 + 25 = 0
g) x2 − 4x + 5 = 0
h) x2 − 4x − 5 = 0
d) x2 + 7 = 0
j) x2 + 6x + 25 = 0
e) 4x2 − 9 = 0
k) 2x2 − 6x − 5 = 0
c) x2 − 7 = 0
f) 4x2 + 9 = 0
i) x2 − 10x + 29 = 0
l) 2x2 + x − 2 = 0
7. Problem solving
The problems in this section can all be converted to algebraic form as quadratic equations, that
can be solved as in the previous sections.
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Problem solving method
Step 1: Carefully read the question until you understand the problem. A rough sketch may be
useful.
Step 2: Decide on the unknown quantity and label it x, say.
Step 3: Use the information given to find an equation which contains x.
Step 4: Solve the equation.
Step 5: Check that any solutions satisfy the equation and are realistic to the problem.
Step 6: Write your answer to the question in sentence form.
Example 10
The sum of a number and its square is 42. Find the number.
Let the number be x. Therefore its square is x2 .
x + x2 = 42
2
∴ x + x − 42 = 0
p
−1 ± 12 − 4 · 1 · (−42)
∴ x =
2
∴ x = −7 or x = 6
So, the number is −7 or 6.
Check: If x = −7,
−7 + (−7)2 = 42 X, and if x = 6,
6 + 62 = 42 X
Example 11
A rectangle has length 5 cm greater than its width. If it has an area of 84 cm2 , find the
dimensions of the rectangle.
If x cm is the width, then (x + 5) cm is the length.
Now area = 84 cm2
∴ x(x + 5) = 84
∴
∴ x2 + 5x = 84
x2 + 5x − 84 = 0
p
52 − 4 · 1 · (−84)
∴ x=
2
∴ x = −12 or x = 7
But x > 0 as lengths are positive quantities, so x = 7. Hence the rectangle is 7 cm by 12 cm.
−5 ±
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Exercises - Set H
1. The sum of a number and its square is 110. Find the number.
2. The product of a number and the number increased by 4 is 117. Find the two possible values
of the number.
3. When 24 is subtracted from the square of a number, the result is five times the original
number. Find the number.
4. The sum of two numbers is 6 and the sum of their squares is 28. Find the exact values of
these numbers.
5. Two numbers differ by 7 and the sum of their squares is 29. Find the numbers.
6. A rectangle has length 4 cm greater than its width. Find its width given that its area is 96
cm2 .
7. A triangle has base 4 m more than its altitude. If its area is 70 m2 , find its altitude.
8. A rectangular enclosure is made from 60 m of fencing. The area enclosed is 216 m2 . Find
the dimensions of the enclosure.
9. A rectangular garden bed was built against an existing brick wall. 24 m of edging was used
to enclosure 60 m2 . Find the dimensions of the garden bed to the nearest cm.
10. ABCD is a rectangle in which AB = 21 cm. The square AXY D is removed and the
remaining rectangle has area 80 cm2 . Find the length of BC.
D
Y
C
A
X
B
11. Sarah has a brother who is 5 years younger than herself and another brother who is 8 years
younger than herself. She observes that the product of her brothers’ ages is equal to the age of
her 40 year old father. How old is Sarah?
12. The numerator of a fraction is 3 less than the denominator. If the numerator is increased
by 6 and the denominator is increased by 5, the fraction is doubled in value. Find the original
fraction.
13. 182 sweets are equally divided among a certain number of children at a party. If the number
of sweets each child receives is one more than the number of children, find the number of
children at the party.
1
14. The sum of a number and its reciprocal is 2 12
. Find the number.
15. The sum of a number and twice its reciprocal is 3 32 . Find the number.
16. In a 42 km marathon race a runner took (n− 4) hours to run the race, running at a constant
speed of (n + 7) km/h. Find the time the runner took to run the race and the speed of the
runner.
✁✃✁✃✁✃✁✃✁✃✁✃✁✃
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