Question

Cube root of 18 by regula falsi method is

Answer 1

Given:

x = 18^(¹/₃)

To find:

Cube root of 18 by Regula Falsi method.

Solution:

x = 18^(¹/₃)

x³ = 18

x³ - 18 = 0

Let f(x) = x³ - 18

Here,

x = 0

f(x) = f(0) = 0³ - 18 = -18

x = 1

f(x) = f(1) = -17

x = 2

f(x) = f(2) = -10

x = 3

f(x) = f(3) = 9

1st Iteration:

Here f(2) = -10 < 0 and f(3) = 9 > 0

∴ Now, Root lies between x₀ = 2 and x₁ = 3

$x_2 = x_0 - f(x_0).\frac{x_1 - x_0}{f(x_1) - f(x_0)}$

$x_2 = 2 - (-10 ). \frac{3 - 2}{9 - (-10)}$

x₂ = 2.5263

f(x₂) = f(2.5263) = 2.52633 - 18 = -1.8764 < 0  

2nd iteration:  

Here f(2.5263) = -1.8764 < 0 and f(3) = 9 > 0

∴ Now, Root lies between x₀ = 2.5263 and x₁ = 3

$x_3 = x_0 - f(x_0).\frac{x_1 - x_0}{f(x_1) - f(x_0)}$

x₃ = 2.608  

f(x₃) = f(2.608) = 2.6083 - 18 = -0.2606 < 0  

3rd iteration:  

Here f(2.608) = -0.2606 < 0 and f(3) = 9 > 0

∴ Now, Root lies between x₀ = 2.608 and x₁ = 3

$x_4 = x_0 - f(x_0).\frac{x_1 - x_0}{f(x_1) - f(x_0)}$

x₄ = 2.6191  

f(x₄) = f(2.6191) = 2.61913 - 18 = -0.0346 < 0  

4th iteration:  

Here f(2.6191) = -0.0346 < 0 and f(3) = 9 > 0  

∴ Now, Root lies between x₀ = 2.6191 and x₁ = 3

$x_5 = x_0 - f(x_0).\frac{x_1 - x_0}{f(x_1) - f(x_0)}$

x₅ = 2.6205  

f(x₅) = f(2.6205) = 2.62053 - 18 = -0.0046 < 0

 

5th iteration:  

Here f(2.6205) = -0.0046 < 0 and f(3) = 9 > 0

∴ Now, Root lies between x₀ = 2.6205 and x₁ = 3

$x_6 = x_0 - f(x_0).\frac{x_1 - x_0}{f(x_1) - f(x_0)}$

x₆ = 2.6207  

f(x₆) = f(2.6207) = 2.62073 - 18 = -0.0006 < 0

 

6th iteration:  

Here f(2.6207) = -0.0006 < 0 and f(3) = 9 > 0  

∴ Now, Root lies between x₀ = 2.6207 and x₁ = 3

$x_7 = x_0 - f(x_0).\frac{x_1 - x_0}{f(x_1) - f(x_0)}$

x₇ = 2.6207  

f(x₇) = f(2.6207) = 2.62073 - 18 = 0 < 0

7th iteration:

Here f(2.6207) = 0 < 0 and f(3) = 9 > 0

∴ Now, Root lies between x₀ = 2.6207 and x₁ = 3

$x_8 = x_0 - f(x_0).\frac{x_1 - x_0}{f(x_1) - f(x_0)}$

x₈ = 2.6207  

f(x₈) = f(2.6207) = 2.62073 - 18 = 0 < 0

Therefore, the cube root of 18 by the Regula Falsi method is 2.6207