Question : If $x=(7+3 \sqrt{5})$, then find the value of $x^2+\frac{1}{x^2}$.
Option 1: $ \frac{580+315 \sqrt{5}}{8}$
Option 2: $\frac{799+328 \sqrt{5}}{8}$
Option 3: $\frac{799+315 \sqrt{5}}{12}$
Option 4: $\frac{799+315 \sqrt{5}}{8}$
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Correct Answer: $\frac{799+315 \sqrt{5}}{8}$
Solution :
Given, $x=(7+3 \sqrt{5})$
Squaring both sides, we get,
$x^2=(7+3 \sqrt{5})^2$
⇒ $x^2=49+9\times 5+2\times 7\times 3\sqrt5$ [As $(a+b)^2=a^2+b^2+2ab$]
⇒ $x^2=49+45+42\sqrt5$
⇒ $x^2=94+42\sqrt5$
And, $\frac{1}{x^2}=\frac{1}{(7+3\sqrt5)^2}=\frac{1}{94+42\sqrt5}$
Rationalising the equation,
⇒ $\frac{1}{x^2}=\frac{1}{94+42\sqrt5}\times\frac{{94-42\sqrt5}}{{94-42\sqrt5}}$
⇒ $\frac{1}{x^2}=\frac{{94-42\sqrt5}}{(94)^2-(42\sqrt5)^2}$
⇒ $\frac{1}{x^2}=\frac{{94-42\sqrt5}}{8836-8820}$
⇒ $\frac{1}{x^2}=\frac{{94-42\sqrt5}}{16}$
⇒ $\frac{1}{x^2}=\frac{{47-21\sqrt5}}{8}$
Now, $x^2+\frac{1}{x^2}$
Putting the value of $x^2$ and $\frac{1}{x^2}$, we get,
= $94+42\sqrt5+\frac{{47-21\sqrt5}}{8}$
= $\frac{752+336\sqrt5+47-21\sqrt5}{8}$
= $\frac{799+315\sqrt5}{8}$
Hence, the correct answer is $\frac{799+315\sqrt5}{8}$.
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