Question

Please solve this question

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}} \\ \\ \\ \sf Let , \: x \: + \: \dfrac{1}{x} \: = \: y \qquad...(1)$




$\underline{\textsf{Squaring both sides , }} \\ \\ \sf \implies \left\{ x \: + \: \dfrac{1}{x} \right\}^2 = \: y^2 \\ \\ \\ \sf \implies x^2 \: + \: \dfrac{1}{x^2} \: + \: 2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \: = \: y^2 \\ \\ \\ \sf \implies x^2 \: + \: \dfrac{1}{x^2} \: + \: 2 \: = \: y^2 \\ \\ \\ \sf \: \: \therefore \: \: x^2 \: + \: \dfrac{1}{x^2} \: = \:y^2 \: - \: 2 \qquad...(2)$





$\underline{\textsf{Given,}} \\ \\ \sf \implies 6\left( x^2 \: + \: \dfrac{1}{x^2} \right) \: - \: 35\left( x \: + \: \dfrac{1}{x} \right) \: + \: 62 \: = \: 0 \\ \\ \underline{\textsf{Plug the value of (1) and (2) , }}$





$\sf \implies 6(y^2 \: - \: 2) \: - \: 35y \: + \: 62 \: = \: 0 \\ \\ \\ \sf \implies 6y^2 \: - \: 12 \: - \: 35y \: + \: 62 \: = \: 0 \\ \\ \\ \sf \implies 6y^2 \: - \: 35y \: + \: 62 \: - \: 12 \: = \: 0 \\ \\ \\ \sf \implies 6y^2 \: - \: 35y \: + \: 50 \: = \: 0 \\ \\ \\ \sf \implies 6y^2 \: - \: 15y \: - \: 20y \: + \: 50 \: = \: 0 \\ \\ \\ \sf \implies 3y(2y \: - \: 5 ) \: - \: 10(2y \: - \: 5 ) \: = \: 0 \\ \\ \\ \sf \implies (2y \: - \: 5)(3y \: - \: 10 ) \: = \: 0$




$\underline{\textsf{By Zero Product Rule : }} \\ \\ \sf \implies 2y \: - \: 5 \: = \: 0 \quad \implies3y \: - \: 10 \: = \: 0 \\ \\ \\ \sf \implies2y = \: 5 \: \: \: \: \: \: \qquad \implies3y \: = \: 10 \\ \\ \\ \sf \: \: \therefore y \: = \: \dfrac{5}{2} \: , \: \dfrac{10}{3}$




$\underline{\textsf{Case 1 : }} \\ \\ \sf \implies x \: + \: \dfrac{1}{x} \: = \: \dfrac{5}{2} \\ \\ \\ \sf \implies \dfrac{x^2 \: + \: 1}{x} \: = \: \dfrac{5}{2} \\ \\ \\ \sf \implies 2(x^2 \: + \: 1) \: = \: 5x \\ \\ \\ \sf \implies 2x^2 \: + \: 1 \: - \: 5x \: = \: 0 \\ \\ \\ \sf \implies 2x^2 \: - \: 5x \: + \: 1 \: = \: 0$




$\underline{\textsf{Here,}} \\ \\ \sf \implies Coefficient \: of \: x^2 (a) \: = \: 2 \\ \\ \sf \implies Coefficient \: of \: x (b) \: = \: -5 \\ \\ \sf \implies Constant \: term(c) \: = \: 1 \\ \\ \\ \underline{\textsf{By Quadratic Formula , }} \\ \\ \sf \implies x \: = \: \dfrac{-b \: \pm \: \sqrt{b^2 \: - \: 4ac}}{2a} \\ \\ \\ \sf \implies x \: = \: \dfrac{-(-5) \: \pm \: \sqrt{(-5)^2 \: - \: 4 \: \cdot \: 2 \: \cdot \: 1}}{2 \: \cdot \: 2}$




$\sf \implies x \: = \: \dfrac{5 \: \pm \: \sqrt{25 \: - \: 8}}{4} \\ \\ \\ \sf \: \: \therefore \: \:x \: = \: \dfrac{5 \: \pm \: \sqrt{17}}{4}$




$\underline{\textsf{Case 2 : }} \\ \\ \sf \implies x \: + \: \dfrac{1}{x} \: = \: \dfrac{10}{3} \\ \\ \\ \sf \implies \dfrac{x^2 \: + \: 1}{x} \: = \: \dfrac{10}{3} \\ \\ \\ \sf \implies 3(x^2 \: + \: 1) \: = \: 10x \\ \\ \\ \sf \implies 3x^2 \: + \:3 \: - \: 10x \: = \: 0 \\ \\ \\ \sf \implies 3x^2 \: - \: 10x \: + \: 3 \: = \: 0 \\ \\ \\ \sf \implies3 {x}^{2} \: - \: 9x \: - \:x \: + \: 3 \: = \: 0 \\ \\ \\ \sf \implies3x( x \: - \: 3) \: - \: (x \: - \: 3) \: = \: 0 \\ \\ \\ \sf \implies(x \: - \: 3)(3x \: - \: 1) \: = \: 0$




$\underline{\textsf{By Zero Product Rule : }} \\ \\ \sf \implies x \: - \: 3\: = \: 0 \quad \implies3x \: - \: 1 \: = \: 0 \\ \\ \\ \sf \implies x \: = \: 3 \: \: \: \: \: \: \qquad \implies3x \: = \: 11\\ \\ \\ \sf \: \: \therefore x \: = \: 3\: , \: \dfrac{1}{3}$




$\textsf{Therefore , } \\ \\ \\ \sf x \: = \: \dfrac{5 \: \pm \: \sqrt{17}}{4} \: , \: 3 \: , \: \dfrac{1}{3}$