Question

find a real root of the equation f(x)=x square - 2x-5=0 using false position method?​

Answer 1

Answer:

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Step-by-step explanation:

Given . Use False-Position method to find a real root of f(x) = x3 - 2x - 5 = 0 correct to three decimal places.

So we have

                      f(x) = x^3 – 2x – 5 = 0

 First we need to find the root so we have

                      f(1) = 1^3 – 2(1) – 5  

                            = 1 – 2 – 5

                            = - 6

                       f(2) = 2^3 – 2(2) – 5

                              = 8 – 4 – 5

                              = - 1

                      f(3) = 3^3 – 2(3) – 5

                              = 27 – 6 – 5

                               = 27 – 11

                               = 16

So the root lies between 2 and 3 (one negative and other positive)

Now using false position method

              Now x1 = 2 and x2 = 3

       f(x1) = f(2) = - 1

      f(x2) = f(3) = 16

So x3 = x1 –  x2 – x1 / f(x2) – f(x1) x f(x1)

     x3 = 2 -  (3 – 2) / 16 – (- 1) x -1

        = 2 -  1 / 17 (-1)

         = 2 + 1/17

          = 35 / 17

          = 2.0588

Now f(2.0588) = (2.0588)^3 – 2(2.0588) – 5

                        = - 0.39105

So the root lies between 2.0588 and 3

  Now x4 = 2.0588 – (3 – 2.0588 / 16 – (- 0.39105)

                 = 2.08125

Next will be x5 = 2.0862

So repeating the process we get x9 = 2.0943 which is correct to three decimal places.