Question:
If $x=3+2 \sqrt{2}$, check whether $x+\frac{1}{x}$ is rational or irrational.
Solution:
$x=3+2 \sqrt{2} \quad \ldots .(1)$
$\Rightarrow \frac{1}{x}=\frac{1}{3+2 \sqrt{2}}$
$\Rightarrow \frac{1}{x}=\frac{1}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$
$\Rightarrow \frac{1}{x}=\frac{3-2 \sqrt{2}}{3^{2}-(2 \sqrt{2})^{2}}$
$\Rightarrow \frac{1}{x}=\frac{3-2 \sqrt{2}}{9-8}$
$\Rightarrow \frac{1}{a}=3-2 \sqrt{2}$......(2)
Adding (1) and (2), we get
$x+\frac{1}{x}=3+2 \sqrt{2}+3-2 \sqrt{2}=6$, which is a rational number
Thus, $x+\frac{1}{x}$ is rational.
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