Question

The root of \({x^4} - 2{x^3} + 5{x^2} - 6 = 0\) in the interval (1,2). Use Newton’s method to approximate the indicated root of the equation correct to six decimal places.

Short answer

The root is \({x_5} = 1.217562\).

Step-by-step solution

Introduction

The Newton’s method about a function is,

\({x_{n + 1}} = {x_n} - \frac{{f({x_n})}}{{f'({x_n})}}\)

f(x) is the function,\(f'({x_n}),{x_n}{\rm{ and }}{{\rm{x}}_{n + 1}}\) are derivation of the function and last two are roots of the function.

Given

\({x^4} - 2{x^3} + 5{x^2} - 6 = 0\)

(1, 2)

The Required initial approximations

\(\begin{aligned}{l}f\left( x \right) &= {x^4} - 2{x^3} + 5{x^2} - 6\\f'\left( x \right) &= 4{x^3} - 6{x^2} + 10x\end{aligned}\)

From newton’s law

\(\begin{aligned}{l}{x_{n + 1}} &= {x_n} - \frac{{x_n^4 - 2x_n^3 + 5x_n^2 - 6}}{{4x_n^3 - 6x_n^2 + 10{x_n}}}\\{x_1} &= 1.5,{\rm{ because interval is }}\left( {1,2} \right)\end{aligned}\)

From the newton’s law,

\(\begin{aligned}{l}{x_2} &= 1.262500,{\rm{ }}\\{x_3} &= 1.218808,{\rm{ }}\\{x_4} &= 1.217563,\\{x_5} &= 1.217562,\\{x_6} &= 1.217562\end{aligned}\)

So, the root is \({x_5} = 1.217562\).