Question

The equation $ {{x}^{3}}-3x+4=0 $ has only one real root. What is its first approximate value as obtained by the method of false position in $ (-3,\,\,-2) $ ?

• A. $ -2.125 $
• B. $ 2.125 $
• C. $ -2.812 $
• D. $ 2.812 $

Answer 1

Answer: (A)

We have, $ {{x}_{2}}={{x}_{0}}-\frac{{{x}_{1}}-{{x}_{0}}}{f({{x}_{1}})-f({{x}_{0}})} $
Here, $ {{x}_{0}}=-3,\,\,{{x}_{1}}=-2,\,\,f(x)={{x}^{3}}-3x+4 $
$ \therefore $ $ f({{x}_{0}})=-27+9+4=-14 $
$ f({{x}_{1}})=-8+6+4=2 $
So, $ {{x}_{2}}=-3-\frac{1}{16}\times (-14) $
$ =-3+\frac{7}{8}=-\frac{17}{8}=-2.125 $