Question

if a = 5+2√6 find the value of a²+1÷a²​

Answer 1

$\huge\text{ANSWER}$

$\huge\text \green{Given}$

$\bf{a = 5 + 2 \sqrt{6} }$

$\huge\text{Explanation}$

$\sf{ \frac{1}{a} = \frac{1}{5 + 2 \sqrt{6}} \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6}} = \frac{5 - 2 \sqrt{6} }{ {(5)}^{2} - {(2 \sqrt{6)} }^{2} } }$

$\sf{ \frac{5 - 2 \sqrt{6} }{25 - 24} = \frac{5 - 2 \sqrt{6} }{1} = 5 - 2 \sqrt{6}}$

$\sf{a + \frac{1}{a} = (5 + 2 \sqrt{6} ) + (5 - 2 \sqrt{6} ) = 10}$

$\huge\text{Now}$

$\sf{( {a}^{2} + \frac{1}{ {a}^{2} })}$

$\sf{ = (a + \frac{1}{a} {)}^{2} - 2a \: \frac{1}{a} }$

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

$= \sf {{(10)}^{2} - 2}$

$\bf{ = 100 - 2 = 98}$

$\huge \boxed{98}$

$\huge\text{THANKS}$

Answer 2

GIVEN :

† a = 5+2√6

TO FIND :

find the value of a²+1÷a²

Identities used :

$\sf: \implies \: {{(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab} \\ \\ \sf:\implies \: {{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab} \\ \\ \sf:\implies \:{ \: (a + b)(a - b) = {a}^{2} - {b}^{2}}$

Given,

a = 5+ 2√6

So,

$\bf{\frac{1}{a} = \frac{1}{5 + 2 \sqrt{6} } }$

On rationalizing the denominator we get,

$\sf : \implies\frac{1}{a} = \frac{1}{5 + 2 \sqrt{6} } \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } \\ \\\sf : \implies \frac{1}{a} = \frac{5 - 2 \sqrt{6} }{ {(5)}^{2} - {(2 \sqrt{6} )}^{2} } \\ \\ \sf : \implies\frac{1}{a} = \frac{5 - 2 \sqrt{6} }{25 - 24} \\ \\ \sf : \implies \frac{1}{a} = 5 - 2 \sqrt{6}$

Now,

$: \implies \sf \: a + \frac{1}{a} = (5 + 2 \sqrt{6} ) + (5 - 2 \sqrt{6} ) \\ \\ : \implies \sf \:a + \frac{1}{a} = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} \\ \\ : \implies \sf \: a + \frac{1}{a} = 10$

Squaring both the sides we get,

$:\implies \sf { (a + \frac{1}{a} )}^{2} = {(10)}^{2} \\ \\ :\implies \sf{a}^{2} + {( \frac{1}{a} )}^{2} + 2 \times a \times \frac{1}{a} = 100 \\ \\ :\implies \sf {a}^{2} + \frac{1}{ {a}^{2} } = 100 - 2 \\ \\ :\implies \sf \: { a }^{2} + \frac{1}{ {a}^{2} } = 98$