Higher Engineering Mathematics, Sixth Edition

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...