Chapter 14

Solving quadratic equations

14.1 Introduction

As stated in Chapter 11, anequationis a statement
that two quantities are equal and to ‘solve an equation’
means ‘to find the value of the unknown’. The value of
the unknown is called therootof the equation.
Aquadratic equationis one in which the highest
power of the unknown quantity is 2. For example,
x^2 − 3 x+ 1 =0 is a quadratic equation.
There are four methods ofsolving quadratic equa-
tions. These are:

(a) by factorization (where possible),

(b) by ‘completing the square’,

(c) by using the ‘quadratic formula’, or

(d) graphically (see Chapter 19).

14.2 Solution of quadratic equations

by factorization

Multiplying out(x+ 1 )(x− 3 )givesx^2 − 3 x+x− 3
i.e.x^2 − 2 x−3. The reverse process of moving from
x^2 − 2 x−3to(x+ 1 )(x− 3 )is calledfactorizing.
If the quadratic expression can be factorized this
provides the simplest method of solving a quadratic
equation.

For example, ifx^2 − 2 x− 3 =0, then, by factorizing
(x+ 1 )(x− 3 )= 0

Hence, either (x+ 1 )=0, i.e. x=− 1

or (x− 3 )=0, i.e. x= 3

Hence,x=−1andx=3 are the roots of the quad-
ratic equationx^2 − 2 x− 3 =0.
The technique of factorizing is often one of trial and
error.

Problem 1. Solve the equationx^2 +x− 6 =0by

factorization

The factors ofx^2 arex andx. These are placed in

brackets: (x )(x )

The factors of−6are+6and−1, or−6and+1, or+ 3

and−2, or−3and+2.

The only combination to give a middle term of+xis

+3and−2,

i.e. x^2 +x− 6 =(x+ 3 )(x− 2 )

The quadratic equationx^2 +x− 6 =0 thus becomes

(x+ 3 )(x− 2 )= 0

Since the only way that this can be true is for either the

first or the second or both factors to be zero, then

either (x+ 3 )= 0 , i.e. x=− 3

or (x− 2 )= 0 , i.e. x= 2

Hence,the roots ofx^2 +x− 6 =0arex=− 3 and

x= 2.

Problem 2. Solve the equationx^2 + 2 x− 8 = 0

by factorization

The factors ofx^2 arex andx. These are placed in

brackets: (x )(x )

The factors of−8are+8and−1, or−8and+1, or+ 4

and−2, or−4and+2.

The only combination to give a middle term of+ 2 xis

+4and−2,

i.e. x^2 + 2 x− 8 =(x+ 4 )(x− 2 )

(Note that the product of the two inner terms( 4 x)added

to the product of the two outer terms(− 2 x)must equal

themiddleterm,+ 2 xin this case.)

DOI: 10.1016/B978-1-85617-697-2.00014-4