Basic Engineering Mathematics, Fifth Edition

Solving quadratic equations 103

The quadratic equationx^2 + 2 x− 8 =0 thus becomes
(x+ 4 )(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,

either (x+ 4 )= 0 , i.e. x=− 4

or (x− 2 )= 0 , i.e. x= 2

Hence,the roots ofx^2 + 2 x− 8 =0arex=− 4 and
x= 2.

Problem 3. Determine the roots of

x^2 − 6 x+ 9 =0 by factorization

x^2 − 6 x+ 9 =(x− 3 )(x− 3 ),i.e.(x− 3 )^2 = 0

The LHS is known as aperfect square.
Hence, x= 3 is the only root of the equation
x^2 − 6 x+ 9 =0.

Problem 4. Solve the equationx^2 − 4 x= 0

Factorizing gives x(x− 4 )= 0

If x(x− 4 )= 0 ,

either x=0orx− 4 = 0

i.e. x=^0 or x=^4

These are the two roots of the given equation. Answers
can always be checked by substitution into the original
equation.

Problem 5. Solve the equationx^2 + 3 x− 4 = 0

Factorizing gives (x− 1 )(x+ 4 )= 0

Hence, either x− 1 =0orx+ 4 = 0

i.e. x= 1 or x=− 4

Problem 6. Determine the roots of 4x^2 − 25 = 0

by factorization

The LHS of 4x^2 − 25 =0isthe difference of two
squares,( 2 x)^2 and( 5 )^2.
By factorizing, 4 x^2 − 25 =( 2 x+ 5 )( 2 x− 5 ),i.e.
( 2 x+ 5 )( 2 x− 5 )= 0

Hence, either ( 2 x+ 5 )= 0 , i.e. x=−

5

2

=− 2. 5

or ( 2 x− 5 )= 0 , i.e. x=

5

2

= 2. 5

Problem 7. Solve the equationx^2 − 5 x+ 6 = 0

Factorizing gives (x− 3 )(x− 2 )= 0

Hence, either x− 3 =0orx− 2 = 0

i.e. x= 3 or x= 2

Problem 8. Solve the equationx^2 = 15 − 2 x

Rearranging gives x^2 + 2 x− 15 = 0

Factorizing gives (x+ 5 )(x− 3 )= 0

Hence, either x+ 5 =0orx− 3 = 0

i.e. x=− 5 or x= 3

Problem 9. Solve the equation 3x^2 − 11 x− 4 = 0

by factorization

The factors of 3x^2 are 3xandx. These are placed in

brackets: ( 3 x )(x )

The factors of−4are−4and+1, or+4and−1, or− 2

and 2.

Remembering that the product of the two inner terms

added to the product of the two outer terms must equal

− 11 x, the only combination to give this is+1and−4,

i.e. 3 x^2 − 11 x− 4 =( 3 x+ 1 )(x− 4 )

The quadratic equation 3x^2 − 11 x− 4 =0 thus

becomes ( 3 x+ 1 )(x− 4 )= 0

Hence, either ( 3 x+ 1 )= 0 , i.e. x=−

1

3

or (x− 4 )= 0 , i.e. x= 4

and both solutions may be checked in the original

equation.

Problem 10. Solve the quadratic equation

4 x^2 + 8 x+ 3 =0 by factorizing

The factors of 4x^2 are 4xandxor 2xand 2x.

The factors of 3 are 3 and 1, or−3and−1.

Remembering that the product of the inner terms added

to the product of the two outer terms must equal+ 8 x,

the only combination that is true (by trial and error) is

( 4 x^2 + 8 x+ 3 )=( 2 x+ 3 )( 2 x+ 1 )