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MATHEMATICS
SUPPORT CENTRE
Title: Quadratic equations.
Target: On completion of this worksheet you should be able to recognise and solve
quadratic equations.
A quadratic equation is an equation where the highest
power of the variable (usually x) is two.
Examples.
2x2 + 3x + 1 = 8 is a quadratic equation.
3x + 2 = 6 and x3 + 2x – 4 = x are not quadratic
equations.
A solution of a quadratic equation is a value of the
variable that makes the equation hold.
E.g. 5 is a solution of the equation x 2 − 2 x − 15 = 0,
since 52-2×5-15=0.
-3 is also a solution of the equation. (Check this for
yourself.)
Often there are two solutions to a quadratic equation, but
sometimes the solutions are identical or don’t exist in
our number system. There will never be more than two
solutions.
In order to solve a quadratic equation we must use an
important property of numbers:
If the product of two values is zero then one of the two
values itself must be zero.
That is if
either
=0
×
=0
or
= 0.
Examples. Solve
a) x2 − 2x – 15 = 0, b) x2 - 2x + 1 = 0,
c) 2x2 − 3x = 0, d) 2x2 + 7x = -3.
a) x 2 − 2 x − 15 = 0
.
⇒ ( x − 5)( x + 3) = 0
⇒ ( x − 5)
× ( x + 3) = 0.
Either x – 5 = 0 or x + 3 = 0.
We now solve these linear equations.
x − 5 = 0 [+5]
x + 3 = 0 [−3]
⇒ x = 5.
⇒ x = −3.
b) x 2 − 2 x + 1 = 0
⇒ ( x − 1)( x − 1) = 0.
⇒ ( x − 1) × ( x − 1) = 0.
Either x − 1 = 0, or x − 1 = 0. Hence x = 1 or
x = 1.
The two solutions are identical.
c) 2 x 2 − 3 x = 0
⇒ x(2 x − 3) = 0
⇒ x × (2 x − 3) = 0.
Either x = 0 or 2x − 3 = 0.
Solving these linear equations gives
3
x = 0 or x = .
2
d ) 2 x 2 + 7 x = −3
[+ 3]
This means that if we can write a quadratic expression
that is equal to zero as the product of two linear
expressions then the value of one of the linear
expressions must be zero. Since we can solve linear
equations we would then be able to solve the quadratic.
⇒ 2x 2 + 7 x + 3 = 0
⇒ (2 x + 1)( x + 3)
⇒ (2 x + 1) × ( x + 3)
Therefore to solve a quadratic we
• Make one side of the equation zero.
• Factorise the quadratic expression.
• Solve the resulting linear equations.
Either 2x + 1 = 0 or x + 3 = 0.
Solving these linear equations gives
1
x = − and x = -3.
2
C. Leech, Coventry University, June 2000.
= 0.
Exercise.
Solve the following quadratic equations.
1. x 2 + 5 x + 6 = 0.
2. x 2 + 12 x + 20 = 0.
3. x 2 + 10 x + 21 = 0.
4. x 2 + 5 x + 4 = 0.
5. x 2 − 2 x − 8 = 0.
6. x 2 = 9 x − 18.
7. 2 x 2 + 13x + 6 = 0.
8. 2 x 2 − 10 x + 12 = 0.
9. 4 x 2 − 16 = 0.
10. 9 y 2 − 25 = 0.
11. 6 x 2 − 4 x = 0.
12. 3x − 8 x 2 = 0.
13. 2 x 2 + 9 x = −4 .
14. 2 x 2 + 12 = −11x.
15. x(2 x − 3) = 5 − 6 x.
16. x(1 − x) + 12 = 2 x( x + 2) − 24.
(Answers: -3, -2; -10, -2; -7, -3; -4, -1; 4, -2; 6, 3;
-1/2, -6; 2, 3; -2, 2; -5/3, 5/3; 0, 2/3; 0, 3/8; -4, -1/2;
4, -3/2; -5/2, 1; -4, 3.)
C. Leech, Coventry University, June 2000.