SOLVING EQUATIONS BY ITERATIVE METHODS 77A
− 3 − 2 − 1 1 220f(x)f(x) = 5 x^2 + 11 x− 17
10− 10− (^17) − 20
0
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=1 andx= 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
Sincef(1) is negative andf(1.25) is positive, a root
lies betweenx=1 andx= 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
Sincef(1) is negative andf(1.125) is positive, a root
lies betweenx=1 andx= 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=1 andx= 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 posi-
tive, a root lies betweenx= 1 .03125 andx= 1 .0625.
Bisecting this interval gives
- 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
positive, a root lies between x= 1 .046875 and
x= 1 .0625.
Bisecting this interval gives - 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 val-
ues are both 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 =0 is 1.05
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...