Basic Engineering Mathematics, Fifth Edition

Graphical solution of equations 159

 10

 8

 6

 4

 2

0.6

y 5 x 2  9 x7.2

0.9 2.4

 1 0 123

2

4

6

8

10

11.25

12

y

x

Figure 19.10

y 2 x^2

y 8

yx

3

 2  10

2

4

6

8

10

A

C

D

B

y

1 1.5 2 x

Figure 19.11

Problem 6. Plot the graph ofy=− 2 x^2 + 3 x+ 6

for values ofxfromx=−2tox=4. Use the

graph to find the roots of the following equations.

(a) − 2 x^2 + 3 x+ 6 =0(b)− 2 x^2 + 3 x+ 2 = 0

(c) − 2 x^2 + 3 x+ 9 =0(d)− 2 x^2 +x+ 5 = 0

A table of values fory=− 2 x^2 + 3 x+6isdrawnupas
shown below.

x − 2 − 1 0 1 2 3 4

y − 8 1 6 7 4 − 3 − 14

A graph of y=− 2 x^2 + 3 x+6isshownin

Figure 19.12.

28

26

24

22

2 1.35 2 1.13

22

2 1.5 G

A B

C D

H

F y^523

y 54

y 52 x 11

y 522 x^213 x 16

x

E

12

1.85 2.63

(^03)
2
4
6
8
y
212 0.5
Figure 19.12
(a) The parabolay=− 2 x^2 + 3 x+6 and the straight
liney=0 intersect atAandB,wherex=− 1. 13
and x= 2. 63 , and these are the roots of the
equation− 2 x^2 + 3 x+ 6 =0.
(b) Comparing y=− 2 x^2 + 3 x+6(1)
with 0 =− 2 x^2 + 3 x+2(2)
shows that, if 4 is added to both sides of
equation (2), the RHS of both equations will
be the same. Hence, 4=− 2 x^2 + 3 x+6. The
solution of this equation is found from the
points of intersection of the line y=4and
the parabola y=− 2 x^2 + 3 x+6; i.e., pointsC
and D in Figure 19.12. Hence, the roots of
− 2 x^2 + 3 x+ 2 =0arex=− 0. 5 andx= 2.
(c) − 2 x^2 + 3 x+ 9 =0 may be rearranged as
− 2 x^2 + 3 x+ 6 =−3 and the solution of
this equation is obtained from the points
of intersection of the line y=−3andthe
parabola y=− 2 x^2 + 3 x+6; i.e., at points E
and F in Figure 19.12. Hence, the roots of
− 2 x^2 + 3 x+ 9 =0arex=− 1. 5 andx= 3.