Higher Engineering Mathematics, Sixth Edition

78 Higher Engineering Mathematics

Problem 1. Use the method of bisection to find

the positive root of the equation 5x^2 + 11 x− 17 = 0

correct to 3 significant figures.

Letf(x)= 5 x^2 + 11 x− 17

then, using functional notation:

f( 0 )=− 17

f(1)= 5 ( 1 )^2 + 11 ( 1 )− 17 =− 1

f(2)= 5 ( 2 )^2 + 11 ( 2 )− 17 =+ 25

Since there is a change of sign from negative

to positive there must be a root of the equation between

x=1andx=2. This is shown graphically in Fig. 9.2.

f(x)

20

10

 4  3  2  1 0 1 2 x

f(x) 5 x^2  11 x 17

 10

 17

 20

Figure 9.2

The method of bisection suggests that the root is at

1 + 2

2

= 1 .5, i.e. the interval between 1 and 2 has been

bisected.

Hence

f(1.5)= 5 ( 1. 5 )^2 + 11 ( 1. 5 )− 17

=+10.75

Sincef( 1 )is negative,f( 1. 5 )is positive, andf( 2 )is

also positive, a root of the equation must lie between

x=1andx= 1 .5, since asign changehas occurred

betweenf( 1 )andf( 1. 5 ).

Bisecting this interval gives

1 + 1. 5

2

i.e. 1.25 as the next

root.

Hence

f(1.25)= 5 ( 1. 25 )^2 + 11 x− 17

=+4.5625

Since f( 1 )is negative andf( 1. 25 )is positive, a root

lies betweenx=1andx= 1 .25.

Bisecting this interval gives

1 + 1. 25

2

i.e. 1.125

Hence

f(1.125)= 5 ( 1. 125 )^2 + 11 ( 1. 125 )− 17

=+1.703125

Since f( 1 )is negative andf( 1. 125 )is positive, a root

lies betweenx=1andx= 1 .125.

Bisecting this interval gives

1 + 1. 125

2

i.e. 1.0625.

Hence

f(1.0625)= 5 ( 1. 0625 )^2 + 11 ( 1. 0625 )− 17

=+0.33203125

Sincef( 1 )is negative andf( 1. 0625 )is positive, a root

lies betweenx=1andx= 1 .0625.

Bisecting this interval gives

1 + 1. 0625

2

i.e. 1.03125.

Hence

f(1.03125)= 5 ( 1. 03125 )^2 + 11 ( 1. 03125 )− 17

=−0.338867...

Sincef( 1. 03125 )is negative andf( 1. 0625 )is positive,

a root lies betweenx= 1 .03125 andx= 1 .0625.

Bisecting this interval gives

1. 03125 + 1. 0625

2

i.e. 1. 046875.

Hence

f(1.046875)= 5 ( 1. 046875 )^2 + 11 ( 1. 046875 )− 17

=−0.0046386...

Since f( 1. 046875 )is negative and f( 1. 0625 )is posi-

tive, a root lies betweenx= 1 .046875 andx= 1 .0625.

Bisecting this interval gives

1. 046875 + 1. 0625

2

i.e.1.0546875.

The last three values obtained for the root are 1.03125,

1.046875 and 1.0546875. The last two values are both