Solving equations by iterative methods 79
1.05, correct to 3 significant figure. We therefore stop
the iterations here.
Thus, correct to 3 significant figures, the positive root
of 5x^2 + 11 x− 17 =0is1.05
Problem 2. Use the bisection method to deter-
mine the positive root of the equationx+ 3 =ex,
correct to 3 decimal places.
Letf(x)=x+ 3 −ex
then, using functional notation:
f(0)= 0 + 3 −e^0 =+ 2
f(1)= 1 + 3 −e^1 =+1.2817...
f(2)= 2 + 3 −e^2 =−2.3890...
Since f( 1 )is positive andf( 2 )is negative, a root lies
betweenx=1andx=2. A sketch off(x)=x+ 3 −ex,
i.e.x+ 3 =exisshowninFig.9.3.
f(x)
1
2
3
4
22 21 0 1 2 x
f(x) 5 x 13
f(x) 5 ex
Figure 9.3
Bisecting the interval betweenx=1andx=2gives
1 + 2
2
i.e. 1.5.
Hence
f(1.5)= 1. 5 + 3 −e^1.^5
=+0.01831...
Sincef( 1. 5 )is positiveandf( 2 )is negative, a root lies
betweenx= 1 .5andx=2.
Bisecting this interval gives
1. 5 + 2
2
i.e. 1.75.
Hence
f(1.75)= 1. 75 + 3 −e^1.^75
=−1.00460...
Sincef( 1. 75 )is negative andf( 1. 5 )is positive, a root
lies betweenx= 1 .75 andx= 1 .5.
Bisecting this interval gives
1. 75 + 1. 5
2
i.e. 1.625.
Hence
f(1.625)= 1. 625 + 3 −e^1.^625
=−0.45341...
Sincef( 1. 625 )is negative andf( 1. 5 )is positive, a root
lies betweenx= 1 .625 andx= 1 .5.
Bisecting this interval gives
1. 625 + 1. 5
2
i.e. 1.5625.
Hence
f(1.5625)= 1. 5625 + 3 −e^1.^5625
=−0.20823...
Since f( 1. 5625 )is negative and f( 1. 5 )is positive, a
root lies betweenx= 1 .5625 andx= 1 .5.
Bisecting this interval gives
1. 5625 + 1. 5
2
i.e. 1. 53125.
Hence
f(1.53125)= 1. 53125 + 3 −e^1.^53125
=−0.09270...
Since f( 1. 53125 )is negative andf( 1. 5 )is positive, a
root lies betweenx= 1 .53125 andx= 1 .5.
Bisecting this interval gives
1. 53125 + 1. 5
2
i.e. 1. 515625.
Hence
f(1.515625)= 1. 515625 + 3 −e^1.^515625
=−0.03664...
Sincef( 1. 515625 )is negative andf( 1. 5 )is positive, a
root lies betweenx= 1 .515625 andx= 1 .5.
Bisecting this interval gives
1. 515625 + 1. 5
2
i.e. 1. 5078125.
Hence
f(1.5078125)= 1. 5078125 + 3 −e^1.^5078125
=−0.009026...