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