Question

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Answer 1

Question:-

If a = 6 - $\sqrt{35}$, find the value of a² + $\frac{1}{a²}$

Solution:-

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

Now, first of all we have to find the value of $\frac{1}{a}$.

So, $\frac{1}{a}$ = $\frac{1}{6 \: - \: \sqrt{35}}$

$\frac{1}{a}$ = $\frac{6 \: + \sqrt{35}}{(6 \: - \: \sqrt{35})(6 \: + \: \sqrt{35})}$

$\frac{1}{a}$ = $\frac{6 \: + \sqrt{35}}{(6)² \: - \: (\sqrt{35})²}$

$\frac{1}{a}$ = $\frac{6 \: + \sqrt{35}}{36 \: - \: 35}$ $(\because \sqrt{35} \: × \: \sqrt{35} \: = \: 35)$

$\frac{1}{a}$ = $\frac{6 \: + \: \sqrt{35}}{1}$

$\frac{1}{a}$ = 6 + $\sqrt{35}$

Now, we have to find the value of a² + $\frac{1}{a²}$

So, a² + $\frac{1}{a²}$ = $(a \: + \frac{1}{a})²$ - 2.a.$\frac{1}{a}$

a² + $\frac{1}{a²}$ = $(6 \: - \sqrt{35} \: + \: 6 \: + \sqrt{35})²$ - 2

a² + $\frac{1}{a²}$ = (12)² - 2

a² + $\frac{1}{a²}$ = 144 - 2

a² + $\frac{1}{a²}$ = 142

The value of a² + $\frac{1}{a²}$ is 142. [Answer]