Question

Solve the given question:

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\underline{\textsf{Given,}} \\ \\ \sf \implies y \: + \: \dfrac{1}{y} \: = \: -2 \\ \\ \sf \implies \dfrac{y^2 \: + \: 1 }{y} \: = \: -2 \\ \\ \sf \implies y^2 \: + \: 1 \: = \: - 2y \\ \\ \sf \implies y^2 \: + \: 2y \: + \: 1 \: = \: 0 \\ \\ \sf \implies (y)^2 \: + \: 2 \cdot y \cdot 1 \: + \: (1)^2 \: = \: 0 \\ \\ \underline{\textsf{Using Algebraic Identity , }} \\ \\ \sf \implies (a^2 \: + \: b^2 \: + \: 2ab ) \: = \: (a \: + \: b)^2 \\ \\ \sf \implies (y \: + \: 1)^2 \: = \: 0 \\ \\ \sf \implies y \: + \: 1 \: = \: 0 \\ \\ \: \: \sf\therefore \: \: y \: = \: -1$



$\underline{\textsf{To Find : }} \\ \\ \sf = y^{50} \: + \: \dfrac{1}{y^{50}} \\ \\ \textsf{Plug the value of y , } \\ \\ \sf = (-1)^{50} \: + \: \dfrac{1}{(-1)^{50}} \\ \\ \sf = 1 \: + \: \dfrac{1}{1} \\ \\ \sf = 1 \: + \: 1 \\ \\ \sf = 2$


$\textsf{If the power of any negative number is} \\ \textsf{even then the result will be positive.}$

Answer 2

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\underline{\textsf{Given,}} \\ \\ \sf \implies y \: + \: \dfrac{1}{y} \: = \: -2 \\ \\ \sf \implies \dfrac{y^2 \: + \: 1 }{y} \: = \: -2 \\ \\ \sf \implies y^2 \: + \: 1 \: = \: - 2y \\ \\ \sf \implies y^2 \: + \: 2y \: + \: 1 \: = \: 0 \\ \\ \sf \implies (y)^2 \: + \: 2 \cdot y \cdot 1 \: + \: (1)^2 \: = \: 0 \\ \\ \underline{\textsf{Using Algebraic Identity , }} \\ \\ \sf \implies (a^2 \: + \: b^2 \: + \: 2ab ) \: = \: (a \: + \: b)^2 \\ \\ \sf \implies (y \: + \: 1)^2 \: = \: 0 \\ \\ \sf \implies y \: + \: 1 \: = \: 0 \\ \\ \: \: \sf\therefore \: \: y \: = \: -1$



$\underline{\textsf{To Find : }} \\ \\ \sf = y^{50} \: + \: \dfrac{1}{y^{50}} \\ \\ \textsf{Plug the value of y , } \\ \\ \sf = (-1)^{50} \: + \: \dfrac{1}{(-1)^{50}} \\ \\ \sf = 1 \: + \: \dfrac{1}{1} \\ \\ \sf = 1 \: + \: 1 \\ \\ \sf = 2$


$\textsf{If the power of any negative number is} \\ \textsf{even then the result will be positive.}$