Question 1:
[10 Marks]
Find a root of the given equation using Newton-Raphson Method. Keep
values correct to four decimal places.
x3  2x2 10x  20  0
Take initial value
x0  1.5
Solution:
f ( x)  x3  2 x 2  10 x  20
Taking Derivative
f '( x)  3x 2  2(2 x)  10
Putting the value of x0
x0  1.5
f '(1.5)  3(1.5)2  4(1.5)  10
f '(1.5)  6.75  6  10
f '(1.5)  22.75
f ( x0 )  (1.5)3  2(1.5)2  10(1.5)  20
f ( x0 )  3.375  4.5  15  20
f ( x0 )  2.875
x1  x0 
f ( x0 )
f '( x0 )
2.875
22.75
x1  1.3737
x1  1.5 
Putting the value of x1 in equation.
f ( x3 )  (1.3737)3  2(1.3737) 2  10(1.3737)  20
f ( x3 )  2.5922  3.7741  13.737  20
f ( x3 )  0.1033
x1  1.3737
f '(1.3737)  3(1.3737) 2  4(1.3737)  10
f '(1.3737)  5.6611  5.4948  10
f '(1.3737)  21.1559
x4  x0 
f ( x2 )
f '( x3 )
x4  1.3737 
x4  1.36882
0.1033
21.1559
Question 2:
[10 Marks]
Find a root of the given equation using three iterations by Bisection
method
x3  x 11  0
Identify the initial bracket
Solution:
f ( x)  x 3  x  11
x2
f (2)  23  2  11
f (2)  8  2  11
f (2)  5
x3
f (3)  33  3  11
f (3)  27  3  11
f (3)  13
Root lies between f (2) and f (3)
now
x0  x1
2
23
x2 
2
x2  2.5
x2 
f (2.5)  (2.5)3  2.5  11
f (2.5)  15.625  2.5  11
f (2.5)  2.125
Root lies between f (2.5) and f (2)
x1  x2
2
2  2.5
x3 
2
x3  2.25
x3 
f (2.25)  (2.25)3  2.25  11
f (2.25)  11.390  2.25  11
f (2.25)  1.859
 x0 , x1  first.
Root lies between f (2.5) and f (2.25)
x2  x3
2
2.25  2.5
x4 
2
x3  2.375
x4 
f (2.375)  (2.375)3  2.375  11
f (2.375)  13.396  2.375  11
f (2.375)  0.0215
Root lies between f (2.25) and f (2.375)