find by regula falsi method the real root of the equation x3 - x2 -2 = 0
Answers
Answer:
Find the root between (2,3) of x3+ - 2x - 5 = 0, by using regular falsi method.
Given
f(x) = x3 - 2 x - 5
f(2) = 23 - 2 (2) - 5 = -1 (negative)
f(3) = 33 - 2 (3) - 5 = 16 (positive)
Let us take a= 2 and b= 3.
The first approximation to root is x1 and is given by
x1 = (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) = 2.0583 - 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
x2 = (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) = 2.0812 - 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
x3 = (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
It is as simlar
Answer:
Explanation:
f(x) = x3 - 2 x - 5
f(2) = 23 - 2 (2) - 5 = -1 (negative)
f(3) = 33 - 2 (3) - 5 = 16 (positive)
Let us take a= 2 and b= 3.
The first approximation to root is x1 and is given by
x1 = (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) = 2.0583 - 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
x2 = (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) = 2.0812 - 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
x3 = (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
It is as simlar