Question

if $x = 3 + \sqrt{8}$then the value of$( {x}^{2} + 1 \div {x}^{2} )$is​

Answer 1

Answer:

Given

$\sf\to x = 3+\sqrt{8}$

To find

$\sf\to x^2+\dfrac{1}{x^2}$

Now we can write as

$\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}$

Now rationalise

$\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}\times\dfrac{3-\sqrt{8}}{3-\sqrt{8}}=\dfrac{3-\sqrt{8}}{9-8} =3-\sqrt{8}$

we get

$\sf\to\dfrac{1}{x} =3-\sqrt{8}$

Using this identities

$\to\sf (a+b)^2= a^2+b^2+2ab$

Now we can write as

$\sf\to\bigg(x+\dfrac{1}{x}\bigg)^2=x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}$

Now put the value of x and 1/x

$\sf\to (3+\sqrt{8}+3-\sqrt{8})^2=x^2+\dfrac{1}{x^2} +2</p><p>$

$\sf\to (3+3)^2=x^2+\dfrac{1}{x^2}$

$\sf\to (6)^2=x^2+\dfrac{1}{x^2} +2$

$\sf\to 36=x^2+\dfrac{1}{x^2} +2$

$\sf\to x^2+\dfrac{1}{x^2} =36-2$

$\sf\to x^2+\dfrac{1}{x^2}$

Answer

$\sf\to x^2+\dfrac{1}{x^2}$

Answer 2

$\begin{gathered}{\bf{Given\::}} \\ \end{gathered}$

$\begin{gathered}\bf{x = 3\: +\: \sqrt{8}} \\ \end{gathered}$

$\begin{gathered} \\ {\bf{To\: Find\::}} \\ \end{gathered}$

$\begin{gathered} \rm{\left( {x}^{2} \:+ \:\dfrac{1}{ {x}^{2}} \right)} \\ \end{gathered}$

$\begin{gathered} \\ {\bf{Solution\::}} \\ \end{gathered}$

First we calculate the value of 'x²'.

$➠ \tt{\:x^2\:=\:\left(3\: +\: \sqrt{8}\right)^2}$

$➠ \tt{\:x^2\:=\:9\: +\:8\:+\:6\sqrt{8}\:}$

$➠ \bf{\:x^2\:=\:17\:+\:6\sqrt{8}\:}$

Now,

We calculate the value of $\begin{gathered} \rm{\left(\dfrac{1}{ {x}^{2}} \right)}. \\ \end{gathered}$

$➠ \begin{gathered} \tt{\:\dfrac{1}{ {(3\:+\:\sqrt{8})}^{2}} } \\ \end{gathered}$

$➠ \begin{gathered} \tt{\:\dfrac{1}{9\:+\:8\:+\:6\sqrt{8}} } \\ \end{gathered}$

$➠ \begin{gathered} \tt{\:\dfrac{1}{17\:+\:6\sqrt{8}} } \\ \end{gathered}$

Rationalize the above equation.

$➠ \begin{gathered} \tt{\:\dfrac{1}{17\:+\:6\sqrt{8}}\times \dfrac{17\:-\:6\sqrt{8}}{17\:-\:6\sqrt{8}} } \\ \end{gathered}$

$➠ \begin{gathered} \tt{\:\dfrac{17\:-\:6\sqrt{8}}{(17\:+\:6\sqrt{8})\:(17\:-\:6\sqrt{8})} } \\ \end{gathered}$

$➠ \begin{gathered} \tt{\:\dfrac{17\:-\:6\sqrt{8}}{(17)^2\:-\:(6\sqrt{8})^2} } \\ \end{gathered}$

$➠ \begin{gathered} \tt{\:\dfrac{17\:-\:6\sqrt{8}}{289\:-\:288} } \\ \end{gathered}$

$➠ \begin{gathered} \tt{\:\dfrac{17\:-\:6\sqrt{8}}{1} } \\ \end{gathered}$

$➠ \bf{\:(17\:-\:6\sqrt{8})}$

Now,

Adding these two results that we calculate.

$\begin{gathered}:\implies\:\rm{x^2\:+\:\dfrac{1}{ {x}^{2}} } \\ \end{gathered}$

$\begin{gathered}:\implies\:\rm{(17\:+\:6\sqrt{8})\:+\:(17\:-\:6\sqrt{8})\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\rm{17\:+\:\cancel{6\sqrt{8}}\:+\:17\:-\:\cancel{6\sqrt{8}}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\rm{17\:+\:17\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\bf{34\:} \\ \end{gathered}$

$\begin{gathered}{\boxed{\bf{\pink{\therefore\:\left( {x}^{2} \:+ \:\dfrac{1}{ {x}^{2}} \right)\:=\:34\:}}}} \\ \end{gathered}$

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⠀⠀⠀⠀: SOME PROPERTIES :

$⠀⠀❶ \bf{(a\:+\:b)^2\:=\:a^2\:+\:b^2\:+\:2ab}$

$⠀⠀❷ \bf{(a\:+\:b)\:(a\:-\:b)\:=\:a^2\:-\:b^2\:}$