Question

x=2+ √5 then (x⁴-1)/x²=?​

Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{x = \sqrt{5} + 2 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: find - \begin{cases} &\sf{\dfrac{ {x}^{4} - 1}{ {x}^{2} } }  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \: :}}}} \end{gathered}$

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \: (x + y)(x - y) = {x}^{2} - {y}^{2} }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy }}}}}} \\ \end{gathered}$

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

Given

$\rm :\implies\:x = \sqrt{5} + 2$

Now,

Consider,

$\rm :\implies\:\dfrac{1}{x} = \dfrac{1}{ \sqrt{5} + 2 }$

$\rm :\implies\:\dfrac{1}{x} = \dfrac{1}{ \sqrt{5} + 2 } \times \dfrac{\sqrt{5} - 2}{\sqrt{5} - 2}$

$\rm :\implies\:\dfrac{1}{x} = \dfrac{\sqrt{5} - 2}{ 5 \: - \: 4 }$

$\rm :\implies\:\dfrac{1}{x} \: = \: \sqrt{5} - 2$

Now,

• We have to find the value of

$\rm :\implies\:\dfrac{ {x}^{4} - 1}{ {x}^{2} }$

$\rm :\implies\:\dfrac{ {x}^{4} }{ {x}^{2} } - \dfrac{1}{ {x}^{2} }$

$\rm :\implies\: {x}^{2} - \dfrac{1}{ {x}^{2} }$

$\rm :\implies\: {(\sqrt{5} + 2)}^{2} - {(\sqrt{5} - 2)}^{2}$

$\rm :\implies\:4 \times \sqrt{5} \times 2$

$\rm :\implies\:8 \sqrt{5}$