QUADRATIC EQUATIONS
Higher Tier
SOLVING QUADRATIC EQUATIONS BY FACTORISING
SOLVING EQUATIONS
Solving any equation means finding the values of x which satisfy the equation.
Example
Solve 2x  6 = 0
Solution
x = 3 satisfies this equation.
SOLVING QUADRATIC EQUATIONS
The solution to a quadratic equation means finding the value or values of x which satisfies the
equation
EXAMPLE
Solve x²  3x + 2 = 0
SOLUTION
x = 1 satisfies the equation since 1²  3(1) + 2 = 0
x = 2 also satisfies the equation since 2²  3(2) + 2 = 0
METHOD1 - GUESSING

In some simple cases, the solutions can be found by trial and error – guessing
EXAMPLE
Solve x² + x  6 = 0 by guessing !
METHOD 2 - FACTORISING
In the first example , the solutions to x²  3x + 2 = 0 were x = 1 and x = 2
x²  3x + 2 can be factorised as (x - 1)(x - 2)
Notice the connection between the solutions and the factors.
The equation x²  3x + 2 = 0
can be written as (x - 1)(x - 2) = 0
Since zero multiplied by anything is zero, we just need to find a value of x which

makes the first bracket zero
or

makes the second bracket zero
In this case
x - 1 = 0 gives x = 1
and
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QUADRATIC EQUATIONS
Higher Tier
x - 2 = 0 gives x = 2
The solutions to the equation x²  3x + 2 = 0 are x = 1 and x = 2
EXAMPLE
Solve x2 + x - 6 = 0
Solution
Factorise
x2 + x - 6 = (x + 3)(x - 2)
x2 + x - 6 = 0 is the same as (x + 3)(x - 2) = 0
x + 3 = 0 gives x = -3
x - 2 = 0 gives x = 2
The solution is x = -3 and x =2
EXAMPLE
Solve 2x2 - 5x - 3 = 0
Solution
Factorise
2x2 - 5x - 3 = (2x + 1)(x - 3)
2x2 - 5x - 3 = 0 is the same as (2x + 1)(x - 3) = 0
2x + 1 = 0 gives x = - 0.5
x - 3 = 0 gives x = 3
The solution is x = -0.5 and x =3
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QUADRATIC EQUATIONS
Higher Tier
EXERCISE 1
Solve by factorising
1.
x² + 3x + 2 = 0
2.
x² + 6x + 5 = 0
3.
x² + 7x + 12 = 0
4.
x² + 8x + 15 = 0
5.
x² + 21x + 20 = 0
6.
x² + 8x + 7 = 0
7.
x² + 8x + 12 = 0
8.
x² + 13x + 12 = 0
9.
x² + 22x + 40 = 0
10.
x² + 11x + 30 = 0
11.
x²  9x + 8 = 0
12.
x²  17x + 30 = 0
13.
x²  5x + 6 = 0
14.
x²  18x + 32 = 0
15.
x²  16x + 63 = 0
16.
x²  x  6 = 0
17.
x² + x  20 = 0
18.
x²  x  12 = 0
19.
x² +2x  15 = 0
20.
x²  8x  20 = 0
21.
x² + 9x + 14 = 0
22.
x²  10x + 21 = 0
23.
x²  10x + 25 = 0
24.
x² + 2x  80 = 0
25.
x²  11x + 30 = 0
26.
x²  12x  28 = 0
27.
x² + 11x + 24 = 0
28.
x²  11x  42 = 0
29.
x²  18x + 32 = 0
30.
x² + 7x  60 = 0
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QUADRATIC EQUATIONS
Higher Tier
EXERCISE 2
Solve by factorising
1.
2x² + 3x + 1 = 0
2.
3x²  5x + 2 = 0
3.
4x² + 7x + 3 = 0
4.
2x²  7x +3 = 0
5.
3x² + 13x + 4 = 0
6.
3x²  8x + 4 = 0
7.
2x² + 9x + 4 = 0
8.
5x²  17x + 6 = 0
9.
2x² + 11x + 12 = 0
10.
7x²  29x + 4 = 0
11.
2x²  3x  2 = 0
12.
3x² + x  4 = 0
quadraticequations
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QUADRATIC EQUATIONS
Higher Tier
USING THE FORMULA
When a quadratic equation cannot easily be solved by factorising,
can be solved by using the formula
Examples
1.
Solve x² +10x + 1 = 0
Solution
Compare with ax² + bx + c = 0
a = 1, b = 10, c = 1
x
b  b2  4ac
2a

10  102  4  1  1
2 1

10  100  4
2
10  96
2
10  9.8

2
10  9.8 0.2
x

 0.1
2
2
or
10  9.8 19.8
x

 9.9
2
2

x = 01 or x = 99 (2 d.p)
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QUADRATIC EQUATIONS
2.
Higher Tier
Solve 2x² + 10x  3 = 0
Solution
a = 2, b = 10, c =  3
x
b  b2  4ac
2a

10  102  4  2  (3)
22

10  100  24
4
10  124
4
10  11.14

4
10  11.14 1.14
x

 0.28
4
4
or
10  11.14 21.14
x

 5.28
4
4

x = 0.28 or 5.28 (2 d.p.)
3.
Solve 3x² = 9x  2
Solution
First re-arrange this equation into the form ax² + bx + c = 0
3x²  9x + 2 = 0
x
 a = 3, b = 9, c = 2
b  b2  4ac
2a

(9)  (9)2  4  3  2
23

9  81  24
6
9  57
6
9  7.55

6
9  7.55 16.55
x

 2.76
6
6
or
9  7.55 1.45
x

 0.24
6
6

x = 2.76 or 0.24 (2 d.p.)
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QUADRATIC EQUATIONS
Higher Tier
EXERCISE 3
Solve the following equations. Give your answers correct to 1 decimal place.
1.
x² + 4x  6 = 0
2.
x² + 2x  4 = 0
3.
5x² + x  4 = 0
4.
x² + 2x + 4 = 0
5.
x² + 3x  1 = 0
6.
2x²  x  7 = 0
7.
3x² = 5x  2
8.
x² = 1  4x
9.
x²  5x + 2 =0
10.
2x² + 3x  1 = 0
Solve the following equations. Give your answers correct to 2 decimal places.
11.
5x²  4x  2 = 0
12.
3x² + x  1 = 0
13.
4x²  x = 2
14.
x² = 4x  3
15.
2  3x  x² = 0
16.
5x² + 7x = 4
17.
6x²  5 = 0
18.
x(x + 2) = 5
19.
(x  2)(x + 1) = 3
20.
x² + 9x + 12 = 0
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QUADRATIC EQUATIONS
Higher Tier
GRAPHICAL SOLUTION

One graph can be used to solve many equations
Example
Draw the graph of y = x² for values of x between 4 and 4.
Use this graph to solve the following equations
a)
x² = 4
b)
x² = 0
c)
x² = 6
d)
x²  9 = 0
e)
2x²  4 = 0
Step 1:
Step 2:
Table of Values for y = x²
X
4
3
2
1
0
1
2
3
4
Y
16
9
4
1
0
1
4
9
16
Draw the axes
Plot the points
Join with a smooth curve
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QUADRATIC EQUATIONS
a) x² = 4
Higher Tier
Draw the horizontal line y = 4
Read off the x - values
x² = 4 when x = 2 and x = 2
b) x² = 0
The horizontal line y = 0 is the x- axis.
x² = 0 when x = 0
c) x² = 6
Draw the horizontal line y = 6
Read off the x - values
x² = 6 when x = 2.45 and x = 2.45
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QUADRATIC EQUATIONS
d) x²  9 = 0
Higher Tier
Rearrange to x² = 9
Draw the horizontal line y = 9
Read off the x – values
x²  9 = 0 when x = 3 and x = 3
e) 2x²  4 = 0
Rearrange to 2x² = 4
And again to x² = 2
Draw the horizontal line y = 2
Read off the x – values
2x²  4 = 0 when x = 1.4 and x = 1.4
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QUADRATIC EQUATIONS
Higher Tier
Example
Draw the graph of y = x²  2x  15 in the range –5  x  6
Use your graph to solve the equations
a)
x²  2x  15 = 0
b)
x² 2x  15 = 10
c)
x²  2x = 14
d)
x²  2x  6 = 0
Step 1:
Step 2:
Table of Values
x
5
4
3
2
1
0
1
2
3
4
5
6
y
20
9
0
7
12
15
16
15
12
7
0
9
Draw the axes
Plot the points
Join with a smooth curve
a) x²  2x  15 = 0
The line y = 0 is the x – axis
y = x²  2x  15 cuts the x-axis when x = 3 and 5
x²  2x  15 = 0 when x = 3 and 5
b) x²  2x  15 = 10
Draw the line y = 10
Read off the x-values
x²  2x  15 = 10 when x = 4.1 and 6.1
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QUADRATIC EQUATIONS
*c) x²  2x = 14
Higher Tier
Rearrange
x²  2x = 14 so that the left hand side is x²  2x  15
x²  2x  15 = 14  15
x²  2x  15 = 1
Draw the horizontal line y = 1
Read off the x – values
x²  2x  15 = 1 when x = -2.9 and 4.9
*d) x²  2x  6 = 0
Rearrange
x²  2x  6 = 0 so that the left hand side is x²  2x  15
x²  2x  6 = 0
x²  2x = 6
x²  2x  15 = 6  15
x²  2x  15 =  9
Draw the horizontal line y = 9
Read off the x – values
x²  2x  6 = 0 when x = -1.6 and 3.6
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QUADRATIC EQUATIONS
Higher Tier
PROBLEM SOLVING

Quadratic equations are often used to solve practical problems.

The method is
o
read the question carefully
o
form a quadratic equation
o
solve this quadratic equation either by factorising or by using the quadratic formula.
Example 1
The sum of the squares of two consecutive whole numbers is 221. Find the whole numbers.
Solution
Let one of the numbers be n
The square of n is n²
The next number is (n + 1)
The square of (n + 1) is (n + 1)²
The equation is
(n + 1)² + n² = 221
Expanding
n² + 2n + 1 + n² = 221
Simplify
2n² + 2n + 1 = 221
2n² + 2n  220 = 0
n² + n  110 = 0
(n + 11)(n  10) = 0
Solve
n =  11 or n = 10
If n = 11, the next number is 10
If n = 10, the next number is 11
There are two answers

11 and 10 or 10 and 11
In some problems, one of the solutions may not be an acceptable answer.
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QUADRATIC EQUATIONS
Higher Tier
Example 2
A rectangle is 1m longer than it is wide. If the area of the rectangle is 20m², find its
dimensions.
Solution
Let the width be x, then the length will be (x + 1)
Area of rectangle = x(x + 1) = x² + x
The equation is x² + x = 20
x² + x  20 = 0
(x + 5)(x  4) = 0
x = 5 or x = 4
Clearly lengths cannot be negative, so x  5
Width = 4 m and the length = 5 m
GCSE Question
ABC is a triangle with a right angle at B.
The length of its three sides are:
x cm, (x + 2) cm and (x + 4) cm.
a) Use Pythagoras’ Theorem to show that x
satisfies the equation
x²  4x  12 = 0
[3]
(x + 4)² = (x + 2)² + x²
b) Solve the equation x²  4x  12 = 0 [2]
(x  6)(x + 2) = 0
x² + 8x + 16 = x² + 4x + 4 + x²
x²  4x  12 = 0
x = 6 or x = 2
c) Use your solutions in (b) to write down the
lengths of the sides of the triangle. [1]
Lengths cannot be negative so x  2
x = 6 cm
AB = 6 cm, BC = 8 cm, AC = 10 cm
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