Question
If $x=3+2 \sqrt{2}$, find the value of $\left(x^2+\frac{1}{x^2}\right)$.
✓
Answer
Given, $x=3+2 \sqrt{2}$
$\therefore \frac{1}{x}=\frac{1}{(3+2 \sqrt{2})}$
$=\frac{1}{(3+2 \sqrt{2})} \times \frac{(3-2 \sqrt{2})}{(3-2 \sqrt{2})}$
$=\frac{(3-2 \sqrt{2})}{(3)^2-(2 \sqrt{2})^2}$
$=\frac{(3-2 \sqrt{2})}{(9-8)}$
$=3-2 \sqrt{2}$
$\therefore x+\frac{1}{x}=(3+2 \sqrt{2})+(3-2 \sqrt{2})$
$x+\frac{1}{x}=6$
$\Rightarrow\left(x+\frac{1}{x}\right)^2=6^2=36$
$\Rightarrow x^2+\frac{1}{x^2}+2 \times x \times \frac{1}{x}=36$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)+2=36$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)=36-2=34$
Hence, $\left(x^2+\frac{1}{x^2}\right)=34$
$\therefore \frac{1}{x}=\frac{1}{(3+2 \sqrt{2})}$
$=\frac{1}{(3+2 \sqrt{2})} \times \frac{(3-2 \sqrt{2})}{(3-2 \sqrt{2})}$
$=\frac{(3-2 \sqrt{2})}{(3)^2-(2 \sqrt{2})^2}$
$=\frac{(3-2 \sqrt{2})}{(9-8)}$
$=3-2 \sqrt{2}$
$\therefore x+\frac{1}{x}=(3+2 \sqrt{2})+(3-2 \sqrt{2})$
$x+\frac{1}{x}=6$
$\Rightarrow\left(x+\frac{1}{x}\right)^2=6^2=36$
$\Rightarrow x^2+\frac{1}{x^2}+2 \times x \times \frac{1}{x}=36$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)+2=36$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)=36-2=34$
Hence, $\left(x^2+\frac{1}{x^2}\right)=34$
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