Question
If $\text{x}=\sqrt{13}+2\sqrt{3},$ find the value of $\text{x}-\frac{1}{\text{x}}.$
✓
Answer
$\text{x}=\sqrt{13}+2\sqrt{3}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt{13}+2\sqrt{3}}$
$=\frac{1}{\sqrt{13}+2\sqrt{3}}\times\frac{\sqrt{13}-2\sqrt{3}}{\sqrt{13}-2\sqrt{3}}$
$=\frac{\sqrt{13}-2\sqrt{3}}{\big(\sqrt{13}\big)^2-\big(2\sqrt{3}\big)^2}$
$=\frac{\sqrt{13}-2\sqrt{3}}{13-12}$
$=\sqrt{13}-2\sqrt{3}$
$\therefore \ \text{x}-\frac{1}{\text{x}}=\sqrt{13}+2\sqrt{3}-\sqrt{13}+2\sqrt{3}=4\sqrt{3}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt{13}+2\sqrt{3}}$
$=\frac{1}{\sqrt{13}+2\sqrt{3}}\times\frac{\sqrt{13}-2\sqrt{3}}{\sqrt{13}-2\sqrt{3}}$
$=\frac{\sqrt{13}-2\sqrt{3}}{\big(\sqrt{13}\big)^2-\big(2\sqrt{3}\big)^2}$
$=\frac{\sqrt{13}-2\sqrt{3}}{13-12}$
$=\sqrt{13}-2\sqrt{3}$
$\therefore \ \text{x}-\frac{1}{\text{x}}=\sqrt{13}+2\sqrt{3}-\sqrt{13}+2\sqrt{3}=4\sqrt{3}$
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