Solve the equation: X3- 5X-5 = 0. This equation cannot be factorised nor solved algebraically therefore it must be solved through interval bisection.

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Interval Bisection.

Ben Ward

Solve the equation:  X3– 5X-5 = 0. This equation cannot be factorised nor solved algebraically therefore it must be solved through interval bisection.

To calculate the approximate position of the roots and the number of roots I used Omnigraph to sketch the graph (below shows where the graph crosses the x-axis).

This is the root I want to find and shows how the solution lies between 2.6 and 2.7

[f (X) = X3 – 5X - 5]

X0 = 2.6   f (X0) =  -0.424

X1 = 2.7   f (X1) = + 1.183

As f(X0) < 0 and f(X1) > 0 it shows that the solution must lie between X0 and X1, thus creating an error bound  2.6 < X < 2.7.

In order to reduce the error bound and ultimately find the solution I set-up a table of values for  X and f(X) using Excel. The spread sheet was used to speed up calculations and therefore ultimately identify the solution to the equation.

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After 16 literations it gives the answer of 2.627 to 3 d.p or 2.627 +/- 0.0005. Therefore it can be said that that 2.627<X<2.628.

  • As the sign of (X) when its equal to 2.627 is negative

            i.e. 2.627 3 – 5*2.627  - 5 =  -0.00573

  • And the sign of (X) when it equals 2.628 is positive

            i.e. 2.628 3 – 5*2.628  - 5 =  -0.00998

To give an example of an equation ...

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