Basic Engineering Mathematics, Fifth Edition

158 Basic Engineering Mathematics

Lety= 4 x^2 + 4 x−15. A table of values is drawn up as

shown below.

x − 3 − 2 − 1 0 1 2

y= 4 x^2 + 4 x− 15 9 − 7 − 15 − 15 − 7 9

12 y 54 x 214 x 215

y

8

4

24

28

212

216

2 0.5

1.5

0 12 x

A B

2 2.5

23 22 21

Figure 19.8

A graph ofy= 4 x^2 + 4 x−15 is shown in Figure 19.8.

The onlypointswherey= 4 x^2 + 4 x−15 andy=0are

the points markedAandB. This occurs atx=− 2. 5

andx= 1. 5 and these are the solutions of the quadratic

equation 4x^2 + 4 x− 15 =0.

By substitutingx=− 2 .5andx= 1 .5 into the original

equation the solutions may be checked.

The curve has a turning point at(− 0. 5 ,− 16 )and the

nature of the point is aminimum.

An alternative graphical method of solving

4 x^2 + 4 x− 15 =0 is to rearrange the equation as

4 x^2 =− 4 x+15 and then plot two separate graphs

−in this case, y= 4 x^2 and y=− 4 x+15. Their

points of intersection give the roots of the equation

4 x^2 =− 4 x+15, i.e. 4x^2 + 4 x− 15 =0. This is shown

in Figure 19.9, where the roots are x=− 2 .5and

x= 1 .5, as before.

Problem 4. Solve graphically the quadratic

equation− 5 x^2 + 9 x+ 7. 2 =0 given that the

solutions lie betweenx=−1andx=3. Determine

also the co-ordinates of the turning point and state

its nature

y 524 x 115

30

25

20

15

10

5

y

22 21 0 1 23 x

2 2.5 1.5

23

y 54 x^2

Figure 19.9

Lety=− 5 x^2 + 9 x+ 7 .2. A tableof values is drawn up

as shown below.

x − 1 − 0. 5 0 1

y=− 5 x^2 + 9 x+ 7. 2 − 6. 8 1.45 7.2 11.2

x 2 2.5 3

y=− 5 x^2 + 9 x+ 7. 2 5.2 − 1. 55 − 10. 8

A graph ofy=− 5 x^2 + 9 x+ 7 .2 is shown plotted in

Figure 19.10. The graph crosses thex-axis (i.e. where

y=0) atx=− 0. 6 andx= 2. 4 and these are the solu-

tions of the quadratic equation− 5 x^2 + 9 x+ 7. 2 =0.

The turning point is amaximum, having co-ordinates

( 0. 9 , 11. 25 ).

Problem 5. Plot a graph ofy= 2 x^2 and hence

solve the equations

(a) 2x^2 − 8 =0(b)2x^2 −x− 3 = 0

A graph ofy= 2 x^2 is shown in Figure 19.11.

(a) Rearranging 2x^2 − 8 =0gives2x^2 =8andthe

solutionofthisequationisobtainedfromthepoints

of intersection ofy= 2 x^2 andy=8; i.e., at co-

ordinates(− 2 , 8 )and( 2 , 8 ),shownasAandB,

respectively, in Figure 19.11. Hence, the solutions

of 2x^2 − 8 =0arex=− 2 andx=+ 2.

(b) Rearranging 2x^2 −x− 3 =0gives2x^2 =x+ 3

and the solution of this equation is obtained

from the points of intersection ofy= 2 x^2 and

y=x+3; i.e., atCandDin Figure 19.11. Hence,

the solutions of 2x^2 −x− 3 =0arex=− 1 and

x= 1. 5