Question

(x²+1/x²)-5(x-1/x)+2=0​

Answer 1

Answer:

$\sf{The \ roots \ of \ the \ equation \ are}$

$\sf{\dfrac{5-\sqrt{33}}{4} \ and \ \dfrac{5+\sqrt{33}}{4}.}$

Given:

$\sf{\leadsto{\dfrac{x^{2}+1}{x^{2}}-5(\dfrac{x-1}{x})+2=0}}$

To find:

$\sf{Roots \ of \ the \ equation.}$

Solution:

$\sf{\leadsto{\dfrac{x^{2}+1}{x^{2}}-5(\dfrac{x-1}{x})+2=0}}$

$\sf{Multiply \ throughout \ by \ x^{2} \ we \ get}$

$\sf{\leadsto{x^{2}+1-5x(x-1)+2=0}}$

$\sf{\leadsto{x^{2}+1-5x^{2}+5x+2x^{2}=0}}$

$\sf{\leadsto{-2x^{2}+5x+1=0}}$

$\sf{Here, \ a=-2, \ b=5 \ and \ c=1}$

$\sf{\therefore{b^{2}-4ac=5^{2}-4(-2)(1)}}$

$\sf{\therefore{b^{2}-4ac=25+8}}$

$\sf{\therefore{b^{2}-4ac=33}}$

$\sf{By \ quadratic \ formula}$

$\sf{x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

$\sf{\leadsto{x=\dfrac{-5\pm\sqrt{33}}{-4}}}$

$\sf{\leadsto{x=\dfrac{5-\sqrt{33}}{4} \ or \ \dfrac{5+\sqrt{33}}{4}}}$

$\sf\purple{\tt{\therefore{The \ roots \ of \ the \ equation \ are}}}$

$\sf\purple{\tt{\dfrac{5-\sqrt{33}}{4} \ and \ \dfrac{5+\sqrt{33}}{4}.}}$

Answer 2

Answer:

The roots of the equation are $\sf{\dfrac{5-\sqrt{− 33}}{4} \ and \ \dfrac{5+\sqrt33}{4}}$

Hope it helps you.