78 Higher Engineering Mathematics
Problem 1. Use the method of bisection to find
the positive root of the equation 5x 2 + 11x − 17 =0
correct to 3 significant figures.
Let f (x) = 5x 2 + 11x − 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 = 1 and x = 2. This is shown graphically in Fig. 9.2.
f(x)
20
10
0
Ϫ1
Ϫ2
Ϫ3
Ϫ4
1
2
x
f (x) ϭ 5x 2 ϩ 11x Ϫ 17
Ϫ10
Ϫ17
Ϫ20
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
Since f (1) is negative, f (1.5) is positive, and f (2) is
also positive, a root of the equation must lie between
x = 1 and x = 1.5, since a sign change has occurred
between f (1) and f (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
+ 11x − 17
= +4.5625
Since f (1) is negative and f (1.25) is positive, a root
lies between x = 1 and x = 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 and f (1.125) is positive, a root
lies between x = 1 and x = 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
Since f (1) is negative and f (1.0625) is positive, a root
lies between x = 1 and x = 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 . . .
Since f (1.03125) is negative and f (1.0625) is positive,
a root lies between x = 1.03125 and x = 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 positive, a root lies between x = 1.046875 and x = 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
Problem 1. Use the method of bisection to find
the positive root of the equation 5x 2 + 11x − 17 =0
correct to 3 significant figures.
Let f (x) = 5x 2 + 11x − 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 = 1 and x = 2. This is shown graphically in Fig. 9.2.
f(x)
20
10
0
Ϫ1
Ϫ2
Ϫ3
Ϫ4
1
2
x
f (x) ϭ 5x 2 ϩ 11x Ϫ 17
Ϫ10
Ϫ17
Ϫ20
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
Since f (1) is negative, f (1.5) is positive, and f (2) is
also positive, a root of the equation must lie between
x = 1 and x = 1.5, since a sign change has occurred
between f (1) and f (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
+ 11x − 17
= +4.5625
Since f (1) is negative and f (1.25) is positive, a root
lies between x = 1 and x = 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 and f (1.125) is positive, a root
lies between x = 1 and x = 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
Since f (1) is negative and f (1.0625) is positive, a root
lies between x = 1 and x = 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 . . .
Since f (1.03125) is negative and f (1.0625) is positive,
a root lies between x = 1.03125 and x = 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 positive, a root lies between x = 1.046875 and x = 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