Question

If $\sf x= 9+4\sqrt{5}$ then find

$\sf x{}^{2}+\frac{1}{x{}^{2}}$​

Answer 1

Answer:

begin{equation}

\nonumber f_{XYZ}(x,y,z) = \left\{

\begin{array}{l l}

\frac{1}{3}(x+2y+3z) & \quad 0 \leq x,y,z \leq 1 \\

& \quad \\

0 & \quad \text{otherwise}

\end{array} \right.

\end{equation}

Answer 2

To Find :-

• Value of $\sf x{}^{2}+\frac{1}{x{}^{2}}$

Solution :-

Given,

• $\sf x= 9+4\sqrt{5}$

Here, we need to find the value of $\sf \frac{1}{x}$

$\sf:\implies \frac{1}{9+4\sqrt{5}}$

$\sf:\implies \frac{1}{9+4\sqrt{5}}\times \frac{9-4\sqrt{5}}{9-4\sqrt{5}}$

$\sf:\implies \frac{9-4\sqrt{5}}{(9){}^{2}-(4\sqrt{5}){}^{2}}$

$\sf:\implies \frac{9-4\sqrt{5}}{81-80}$

$\sf:\implies \frac{9-4\sqrt{5}}{1}$

$\sf:\implies 9-4\sqrt{5}$

$\sf Hence,$

• The value of $\sf\frac{1}{x}= 9-4\sqrt{5}$

____

Now,

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

$\sf:\implies 9+4\sqrt{5}+9-4\sqrt{5}$

$\sf:\implies 9+9+4\sqrt{5}-4\sqrt{5}$

$\sf:\implies 18$

Here, $\sf x+\frac{1}{x}=18$

____

Again,

$\sf:\implies x{}^{2}+\frac{1}{x{}^{2}}$

$\sf:\implies (x+\frac{1}{x}){}^{2}=x{}^{2}+\frac{1}{x{}^{2}}+2$

$\sf:\implies x{}^{2}+\frac{1}{x{}^{2}}+2=18$

$\sf\implies x{}^{2}+\frac{1}{x{}^{2}}=18-2$

$\sf\implies x{}^{2}+\frac{1}{x{}^{2}}=16$

$\boxed {\sf {\purple {The \ required \ Value \ is\ 16.}}}$

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