Question

The value of $\sqrt{12}$ correct to 3 decimal places by Newton -Raphson method is given by

• A. 3.463
• B. 3.462
• C. 3.467
• D. None of these

Answer 1

Answer: (A)

3.463

Answer 2

Let $x = \sqrt{12} \Rightarrow x^{2} - 12 = 0$ $\Rightarrow f(x) = x^{2} - 12,f'(x) = 2x$

$\therefore$ $f(3) = - 3$ i.e. –ive and $f(4) = + 4$ i.e. +ive

Hence roots lie between 3 and 4. $\because$ $|f(3)|\mspace{6mu} < \mspace{6mu}|f(4)|$, $\therefore\mspace{6mu}\mspace{6mu}\mspace{6mu} x_{0} = 3$

First iteration ,$x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})} = 3 - \frac{- 3}{6} = 3.5$

Now second iteration, $f(3.5) = 0.25,f'(3.5) = 7$,

$x_{2} = 3.5 - \frac{0.25}{7} = 3.463$