Question

2(x^2+1/x^2)-9(x+1/x)+14=0 find its roots​

Answer 1

Answer:

I hope this answer would be helpful for you

Follow me if you are satisfied with answer

Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

$\mathsf{Given :\;\;2\bigg(x^2 + \dfrac{1}{x^2}\bigg) - 9\bigg(x + \dfrac{1}{x}\bigg) + 14 = 0}$

$\mathsf{\bigg(x^2 + \dfrac{1}{x^2}\bigg)\;can\;be\;written\;as :}$

$\mathsf{\bigstar\;\; \bigg(x^2 + \dfrac{1}{x^2}\bigg) = \bigg(x + \dfrac{1}{x}\bigg)^2 - 2}$

$\mathsf{\implies 2\bigg[\bigg(x + \dfrac{1}{x}\bigg)^2 - 2\bigg] - 9\bigg(x + \dfrac{1}{x}\bigg) + 14 = 0}$

$\mathsf{Let\;us\;take : \bigg(x + \dfrac{1}{x}\bigg) = P}$

$\mathsf{\implies 2(P^2 - 2) - 9P + 14 = 0}$

$\mathsf{\implies 2P^2 - 4 - 9P + 14 = 0}$

$\mathsf{\implies 2P^2 - 9P + 10 = 0}$

$\mathsf{\implies 2P^2 - 4P - 5P + 10 = 0}$

$\mathsf{\implies 2P(P - 2) - 5(P - 2) = 0}$

$\mathsf{\implies (P - 2)(2P - 5) = 0}$

$\mathsf{\implies P = 2\;\;(or)\;\; P = \dfrac{5}{2}}$

$\mathsf{But\;we\;took : \bigg(x + \dfrac{1}{x}\bigg) = P}$

$\underline{\textsf{Case : 1}}$

$\mathsf{Consider : \bigg(x + \dfrac{1}{x}\bigg) = 2}$

$\mathsf{\implies \bigg(\dfrac{x^2 + 1}{x}\bigg) = 2}$

$\mathsf{\implies x^2 + 1 = 2x}$

$\mathsf{\implies x^2 - 2x + 1 = 0}$

$\mathsf{\implies (x - 1)^2 = 0}$

$\mathsf{\implies (x - 1) = 0}$

$\mathsf{\implies x = 1}$

$\underline{\textsf{Case : 2}}$

$\mathsf{Consider : \bigg(x + \dfrac{1}{x}\bigg) = \dfrac{2}{5}}$

$\mathsf{\implies \bigg(\dfrac{x^2 + 1}{x}\bigg) = \dfrac{5}{2}}$

$\mathsf{\implies 2(x^2 + 1) = 5x}$

$\mathsf{\implies 2x^2 - 5x + 2 = 0}$

$\mathsf{\implies 2x^2 - 4x - x + 2 = 0}$

$\mathsf{\implies 2x(x - 2) - (x - 2) = 0}$

$\mathsf{\implies (x - 2)(2x - 1) = 0}$

$\mathsf{\implies x = 2\;\;(or)\;\; x = \dfrac{1}{2}}$

$\underline{\textsf{Combining the results of Case : 1 and Case : 2}}$

$\textsf{The Values of x which satisfy the given Equation are :}$

$\mathsf{\bigstar\;\;x = 1}$

$\mathsf{\bigstar\;\;x = 2}$

$\mathsf{\bigstar\;\;x = \dfrac{1}{2}}$