Question

if x²+1÷x²=7,then find the value of x+1÷x​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{\tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 7} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{x + \dfrac{1}{x} }  \end{cases}\end{gathered}\end{gathered}$

$\large\sf{\underbrace{\underline{Formula\ to\ be\ used:-}}}$

$\tt\implies \: \boxed{ \red{ \bf \: {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2} }}$

$\large\underline\purple{\bold{Solution :- }}$

$\sf{\underbrace{\underline{ \green{ \bf \: According\ to\ the\ question:-}}}}$

$: \implies \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 7$

★ On adding 2 on both sides, we get

$: \implies \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 7 + 2$

$: \implies \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 9$

$: \implies \tt \: {(x + \dfrac{1}{x}) }^{2} = 9$

$: \implies \tt \: x + \dfrac{1}{x} = \pm \: 3$