Show that: x^2+1 x^2 =34, if x=3+2 2 - Vidyadip

Question

Show that$: x^2+\frac{1}{x^2}=34$, if $\mathrm{x}=3+2 \sqrt{2}$

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Answer

Given: $x=3+2 \sqrt{2}$
$\frac{1}{x}=\frac{1}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$
$\frac{1}{x}=\frac{3-2 \sqrt{2}}{(3)^2-(2 \sqrt{2})^2}$
$\frac{1}{x}=\frac{3-2 \sqrt{2}}{9-8}$
$\frac{1}{x}=3-2 \sqrt{2}$
Now, $x+\frac{1}{x}=3+2 \sqrt{2}+3-2 \sqrt{2}$
$x+\frac{1}{x}=6$
Squaring on both sides
$\left(x+\frac{1}{x}\right)^2=(6)^2$
$\Rightarrow x^2+\frac{1}{x^2}+2 \times \not x^{\prime} \times \frac{1}{\not x}=36$
$=x^2+\frac{1}{x^2}=36-2$
$=x^2+\frac{1}{x^2}=34$

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