Basic Engineering Mathematics, Fifth Edition

106 Basic Engineering Mathematics

Adding to both sides (half the coefficient ofx)^2 gives

x^2 +

9

2

x+

(

9

4

) 2

=

(

9

4

) 2

− 4

The LHS is now a perfect square. Thus,

(

x+

9

4

) 2

=

81

16

− 4 =

81

16

−

64

16

=

17

16

Taking the square root of both sides gives

x+

9

4

=

√(

17

16

)

=± 1. 031

Hence, x=−

9

4

± 1. 031

i.e.x=− 1. 22 or− 3. 28 , correct to 3 significant figures.

Problem 17. By completing the square, solve the

quadratic equation 4. 6 y^2 + 3. 5 y− 1. 75 =0, correct

to 3 decimal places

4. 6 y^2 + 3. 5 y− 1. 75 = 0

Making the coefficient ofy^2 unity gives

y^2 +

3. 5

4. 6

y−

1. 75

4. 6

= 0

and rearranging gives y^2 +

3. 5

4. 6

y=

1. 75

4. 6

Adding to both sides (half the coefficient ofy)^2 gives

y^2 +

3. 5

4. 6

y+

(

3. 5

9. 2

) 2

=

1. 75

4. 6

+

(

3. 5

9. 2

) 2

The LHS is now a perfect square. Thus,

(

y+

3. 5

9. 2

) 2

= 0. 5251654

Taking the square root of both sides gives

y+

3. 5

9. 2

=

√

0. 5251654 =± 0. 7246830

Hence, y=−

3. 5

9. 2

± 0. 7246830

i.e. y= 0. 344 or− 1. 105

Now try the following Practice Exercise

PracticeExercise 55 Solving quadratic

equations by completing the square

(answers on page 346)

Solve the following equations correct to 3 decimal

places by completing the square.

1. x^2 + 4 x+ 1 =02.2x^2 + 5 x− 4 = 0


2. 3x^2 −x− 5 =04.5x^2 − 8 x+ 2 = 0


3. 4x^2 − 11 x+ 3 =06.2x^2 + 5 x= 2

14.4 Solution of quadratic equations

by formula

Let the general form of a quadratic equation be given

byax^2 +bx+c=0, wherea,bandcare constants.

Dividingax^2 +bx+c=0byagives

x^2 +

b

a

x+

c

a

= 0

Rearranging gives x^2 +

b

a

x=−

c

a

Adding to each side of the equation the square of half

thecoefficient of theterm inxtomake theLHS a perfect

square gives

x^2 +

b

a

x+

(

b

2 a

) 2

=

(

b

2 a

) 2

−

c

a

Rearranging gives

(

x+

b

a

) 2

=

b^2

4 a^2

−

c

a

=

b^2 − 4 ac

4 a^2

Taking the square root of both sides gives

x+

b

2 a

=

√(

b^2 − 4 ac

4 a^2

)

=

±

√

b^2 − 4 ac

2 a

Hence, x=−

b

2 a

±

√

b^2 − 4 ac

2 a

i.e. the quadratic formula is x=

−b±

√

b^2 − 4 ac

2 a

(This method of obtainingthe formula is completingthe

square−as shown in the previous section.)

In summary,

ifax^2 +bx+c= 0 thenx=

−b±

√

b^2 − 4 ac

2 a

This is known as thequadratic formula.