Question
If $\text{x}=3+2\sqrt{2},$ check whether $\text{x}+\frac{1}{\text{x}}$ is rational or irrational.
✓
Answer
$\text{x}=3+2\sqrt{2}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=3+2\sqrt{2}+\frac{1}{3+2\sqrt{2}}$
$=3+2\sqrt{2}+\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{9-8}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{1}$
$=3+2\sqrt{2}+3-2\sqrt{2}$
$=6$ Thus, the given number is rational.
$\Rightarrow\text{x}+\frac{1}{\text{x}}=3+2\sqrt{2}+\frac{1}{3+2\sqrt{2}}$
$=3+2\sqrt{2}+\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{9-8}$
$=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{1}$
$=3+2\sqrt{2}+3-2\sqrt{2}$
$=6$ Thus, the given number is rational.
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