# Method Of False Position

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Since, f(1.525012) f(2) < 0, the root lies in the interval (1.525012, 2).

Since, f(1.529462) f(2) < 0, the root lies in the interval (1.529462, 2).

Since, f(1.531116) f(2) < 0, the root lies in the interval (1.531116, 2).

Since, f(1.531729) f(2) < 0, the root lies in the interval (1.531729, 2).

Now, |x10 – x9 | = | 1.531956 – 1.53179 | ≈ 0.000227 < 0.0005.

The root has been computed correct to three decimal places. The required root can be taken as x ≈ x_{10} = 1.531956. Note that the right end point x = 2 is fixed for all iterations.