Q98 / 170 · Manipal (MET) 2020 · Mathematics

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

  • Correct answer
    A
    $1.871$
  • B
    $1.868$
  • C
    $1.856$
  • D
    None of these

Solution

Given, $f(x)=x^4-x-10$ We assume $x_0=2$ is the approximate root of $f(x)$. Then, $h=-\frac{f\left(x_0\right)}{f\left(x_0\right)}=-\frac{f(2)}{f(2)}$ $\begin{array}{ll}\Rightarrow & h=-\left[\frac{(2)^4-2-10}{4(2)^3-1}\right] \\ \Rightarrow & h=-\left[\frac{16-12}{31}\right]=\frac{-4}{31} \\ \Rightarrow & h=-0.129\end{array}$ $\therefore$ Positive square root of $f(x)$ by Newton-Raphson method, $\begin{aligned} x_1=x_0+h & =2+(-0.129) \\ & =2-0.129=1.871\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions