Question

if x=5+2√6,then find x2+1/x2.​

Answer 1

$\LARGE{\sf{\underline{\underline{Solution:-}}}}$

Given:-

$\sf{{x = 5 + 2\sqrt{6}}}$

To Find:-

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

As we know,

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

So let us firstly find $\sf{{x+}}$$\sf\dfrac{1}{x}$

$\sf{{x = 5 + 2\sqrt{6}}}$

$\implies \sf\dfrac{1}{x}$$=\sf\dfrac{1}{ 5 + 2\sqrt{6}}$

Rationalizing Factor = 5 - 2√6

$\implies \dfrac{1}{x}$$=\sf\dfrac{1}{ 5 + 2\sqrt{6}}$$\times\sf\dfrac{5 - 2\sqrt{6}}{ 5 - 2\sqrt{6}}$

$=\sf\dfrac{5 - 2\sqrt{6}}{(5)^{2}-(2\sqrt{6})^{2}}$

$=\sf\dfrac{5 - 2\sqrt{6}}{25 - 24}$

$=\sf\dfrac{5 - 2\sqrt{6}}{1}$

So now,

$\sf{{x+}}$$\sf\dfrac{1}{x}$

$\sf{{= 5 + 2\sqrt{6}}}$$\sf{{+ 5 - 2\sqrt{6}}}$

$\sf{{= 5 + 5 + 2\sqrt{6} - 2\sqrt{6}}}$

$\sf{{= 10}}$

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

$\implies \sf{{(10)^{2}}}$$\sf{{= x^{2}}}$$+\sf\dfrac{1}{x^{2}}$$\sf{{+2}}$

$\implies \sf{{100}}$$\sf{{= x^{2}}}$$+\sf\dfrac{1}{x^{2}}$$\sf{{+2}}$

$\implies \sf{{100-2}}$$\sf{{= x^{2}}}$$+\sf\dfrac{1}{x^{2}}$

$\implies \sf{{ x^{2}}}$$+\sf\dfrac{1}{x^{2}}$$\sf{{=98}}$

_____________________________________

Therefore,

$\sf{{x^{2}}}$$+\sf\dfrac{1}{x^{2}}$$\sf{{=98}}$

Answer 2

$\large \sf \underline{ \underline{ \red{given : }}}$

$\tt{x = 5 + 2 \sqrt{6} }$

$\large \sf \underline{ \underline{ \red{to \: find : }}}$

$\tt{x}^{2} + \frac{1}{ {x}^{2} }$

$\large \sf \underline{ \underline{ \red{to \: know : }}}$

$\tt \boxed{ \pink{( {a + b) }^{2} = {a}^{2} + {b}^{2} + 2ab }}$

Here ,

• a = x

• b = 1/x

$\large \sf \underline{ \underline{ \red{solution : }}} \:$

Firstly , we will find 1/x.

$\tt{ \frac{1}{x} = \frac{1}{5 + 2 \sqrt{6} } } \\ \\ \ \ \\ \sf{multiplying \: and \: dividing \: by \: (5 - 2 \sqrt{6}) } \\ \\ \\ = \tt\frac{1}{5 + 2 \sqrt{6} } \: \times \: \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } \\ \\ \\ \tt{ = \frac{5 - 2 \sqrt{6} }{ {5}^{2} - ({2 \sqrt{6} )}^{2} } } \\ \\ \\ \tt{ = \frac{5 - 2 \sqrt{6} }{25 - 24} } \\ \\ \\ \implies \boxed{ \tt{ \frac{1}{x} = 5 - 2 \sqrt{6} }}$

━═━═━═━═━═━═━═━═━

$\rm{we \: will \: find \: (x + { \frac{1}{x}) }^{2} }$

• x = 5 + 2√6

• 1/x = 5 - 2√6

$\tt{(x + { \frac{1}{x}) }^{2} } = (5 + \cancel{ 2 \sqrt{6}} + 5 - \cancel{ 2 \sqrt{6}})^{2} \\ \\ \rm (by \: identity)..... \\ \\ \tt{ {x}^{2} +{ \frac{1}{x} }^{2} + 2( \cancel{x})( \frac{1}{ \cancel{x}} )} = {10}^{2} \\ \\ \\ \tt{ {x}^{2} + \frac{1}{x {}^{2} } + 2 = 100 } \\ \\ \\ \boxed{ \tt{ \blue{{x}^{2} + \frac{1}{ {x}^{2} } } = \blue{ 98}}}$