Question

By Newton-Raphson method, the positive root of the equation $x^{4}-x-10=0$ is

• A. 1.871
• B. 1.868
• C. 1.856
• D. None of these

Answer 1

Answer: (A)

Given, $f(x)=x^{4}-x-10$
We assume $x_{0}=2$ is the approximate root of $f(x)$.
Then, $h=-\frac{f\left(x_{0}\right)}{f'\left(x_{0}\right)}=-\frac{f(2)}{f'(2)}$
$\Rightarrow h=-\left[\frac{(2)^{4}-2-10}{4(2)^{3}-1}\right]$
$\Rightarrow h=-\left[\frac{16-12}{31}\right]=\frac{-4}{31}$
$\Rightarrow h=-0.129$
$\therefore $ Positive square root of $f(x)$ by Newton-Raphson method,
$x_{1}=x_{0}+h=2+(-0.129)$
$=2-0.129=1.871$