Question

If a= 6-\sqrt{35} find the value of a^{2} + \frac{1}{a^{2} }


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Answer 1

Answer:

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Answer 2

Given :

$a \: = 6 - \sqrt{35}$

To find :

${a}^{2} + \frac{1}{ {a}^{2} }$

Solution :

${a}^{2} = {(6 - \sqrt{35} )}^{2}$

$= {a}^{2} = 36 - 12 \sqrt{36} + 36$

$= 72 - 12 \sqrt{36}$

$\frac{1}{ {a}^{2} } = \frac{1}{72 - 12 \sqrt{36} }$

Rationalising the denominator,

$\frac{72 + 12 \sqrt{36} }{5184 - 5184} = 0$

$= 72 - 12 \sqrt{36} + 0$

$= 72 - 12 \sqrt{36}$

Hence, the above is the answer.