Question

If x=square root 5 +2,find the value of x power 4+1upon xpower4

Answer 1

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

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

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

Given

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

Consider

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

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

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

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

$:  \implies  \tt \: \boxed{ \red{x \: + \dfrac{1}{x} = 2 \sqrt{5} }}$

● On squaring both sides, we get

$:  \implies  \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 20$

$:  \implies  \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 20$

$:  \implies  \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 20 - 2$

$:  \implies  \tt \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 18$

● On squaring both sides we get

$:  \implies  \tt \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 \times {x}^{2} \times \dfrac{1}{ {x}^{2} } = 324$

$:  \implies  \tt \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 = 324$

$:  \implies  \tt \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 324 - 2$

$:  \implies  \large\boxed {\purple{ \tt \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 322}}$