Question

Solve using false position method by 3 iterations

Answer 1

Given Equation is f(x) = x^3 + 2x^2 - 8 = 0

f(x) = x^3 + 2x^2 - 8.

f(0) = -8.

f(1) = 1^3 + 2 - 8 

     = 3 - 8

     = -5.


f(2) = 2^3 + 2 - 8

      = 8 + 2*2^2 - 8

      = 8 + 8 - 8 

      = 8.


1st Iteration:

f(1) = -5 and f(2) = 8.

Therefore the root lies between x0 = 1 and x1 = 2.

Then x2 = x0 - f(x0) * x1 - x0/f(x1) - f(x0)

              = 1 - (-5) * 2 - 1/8 + 5

              = 1.38462.


Now f(x2) = (1.38462)^3 + 2(1.38462)^2 - 8

                 = -1.5111 < 0.



2nd Iteration:

f(1.38462) = -1.51115 < 0 and f(2) = 8 > 0

    The root lies between x0 = 1.38462 and x1 = 2.

Then x3 = x0 - f(x0) * x1-x0/f(x1) - f(x0)

              = 1.38 - (-1.51) * 2 - 1.38/8 + 1.51
             
             = 1.48239.


f(x3) = f(1.48239) = -0.34753 < 0.



3rd Iteration:

f(1.48239) = -0.34753 < 0 and f(2) = 8 > 0

The root lies between x0 = 1.48239 and x1 = 2.

Then x4 = x0 - f(x0) * x1 - x0/f(x1) - f(x0)

               = 1.48 - (-0.35) * 2 - 1.48/8 + 0.35

               = 1.50394

f(x4) = (1.50394)^3 + 2(1.50394)^2 - 8

        = -0.074.



Hope this helps!