Question 1:
[10 Marks]
Find a root of the given equation using Newton-Raphson Method. Keep
values correct to four decimal places.
x3 + 2 x2 +10 x − 20 = 0
Take initial value
x0 = 1.5
Solution:
f ( x) = x 3 + 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)