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, $h=-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}=-\frac{f(2)}{f^{\prime }(2)}$
$\Rightarrow h=-\left[\frac{(2{)}^4-2-10}{4(2{)}^3-1}\right]$
$\Rightarrow h=-\left[\frac{16-12}{31}\right]=\frac{-4}{31}$
$\Rightarrow h=-0.129$
$\therefore$ Positive square root of $f(x)$ by Newton-Raphson method,
$x_1=x_0+h=2+(-0.129)$
$=2-0.129=1.871$