1. LECTURE 3 OF 4
SOLUTIONS OF NON
-LINEAR EQUATIONS

2. OBJECTIVES
At the end of the lesson, students will be able to
a) find the root by the Newton-Raphson method using the formula

3. 7.2 (II) : Newton-Raphson Method
If x1 is an approximation to a root
of f(x)=0, then a better approximation
x2 is given by
Repeat this process as required

4. Example 1 Solution Using the Newton-Raphson method,
find the solution of f(x)=x + ex near
x=-0.5 to three decimal places.
Solution
Let f(x) = x + ex
So f’(x) = 1 + ex
x1=-0.5

5. The required solution is x = -0.567
(three decimal places)

6. Example 2 Show that the equation x3 + x – 6 = 0
has a root between 1 and 2.
Using the Newton-Raphson method,
with the starting point 1.6 to determine
an approximation to this root, giving
your answer to three significant figures.

7. Solution Let f(x) = x3 + x - 6 f(1) = (1)3 + 1 – 6 = -4 <0 (-ve)
Since f(1) < 0 and f(2) > 0 , so x3 +x – 6 = 0
has a root between x = 1 and x = 2

8. Newton-Raphson Method
Let f(x)= x3 +x – 6
So, f’(x)= 3x2 +1
x1 = 1.6 =

9. So, to three significant figures the
root is 1.63

10. Exercise 1. By taking 0.2 as a first approximation
find the root of the equation
giving your answer to
three significant figures by using the
Newton Raphson method.
(Ans : 0.246)

11. 2. Show that the equation x3 - 2x2 = 4
has a root between 2 and 3.
Using the Newton-Raphson method,
determine an approximation to this
root, giving your answer to three significant figures.
(Ans : 2.59)