The Regula falsi method is an oldest method for computing the real roots of an algebraic equation. This below worksheet help you to understand how to compute the roots of an algebraic equation using Regula falsi method. The practice problems along with this worksheet improve your problem solving capabilities when you try on your own

**Examples:**

Find the root between (2,3) of x^{3}+ - 2x - 5 = 0, by using regular falsi method.

Given

f(x) = x^{3} - 2 x - 5

f(2) = 2^{3} - 2 (2) - 5 = -1 (negative)

f(3) = 3^{3} - 2 (3) - 5 = 16 (positive)

Let us take a= 2 and b= 3.

The first approximation to root is x^{1} and is given by

x_{1} = (a f(a) - b f(b))/(f(b)-f(a))

=(2 f(3)- 3 f(2))/(f(3) - f(2))

=(2 x 16 - 3 (-1))/ (16- (-1))

= (32 + 3)/(16+1) =35/17

= 2.058

Now f(2.058) = 2.058^{3} - 2 x 2.058 - 5

= 8.716 - 4.116 - 5

= - 0.4

The root lies between 2.058 and 3

Taking a = 2.058 and b = 3. we have the second approximation to the root given by

x_{2} = (a f(a) - b f(b))/(f(b)-f(a))

= (2.058 x f(3) - 3 x f(2.058)) /(f(3) - f(2.058))

= (2.058 x 16 -3 x -0.4) / (16 - (-0.4))

= 2.081

Now f(2.081) = 2.081^{2} - 2 x 2.081 - 5

= -0.15

The root lies between 2.081 and 3

Take a = 2.081 and b = 3

The third approximation to the root is given by

x_{3} = (a f(a) - b f(b))/(f(b)-f(a))

= (2.089 X 16 - 3 x (-0.062))/ (16 - (-0.062))

= 2.093

The root is 2.09

**Practice Problems:**

1. Find the approximate value of the real root of *x log _{10} x = 1.2* by regula falsi method

2. Find the root of the

3. Find a root which lies between 1 and 2 of f(x) = x

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