Question
If $\text{a}=2+\sqrt{3},$ then find the value of $\text{a}-\frac{1}{\text{a}}.$
✓
Answer
We have $\text{a}=2+\sqrt{3},$
$\therefore\ \frac{1}{\text{a}}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}$
$=\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}$
$\therefore\ \text{a}-\frac{1}{\text{a}}=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}$
$\therefore\ \frac{1}{\text{a}}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}$
$=\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}$
$\therefore\ \text{a}-\frac{1}{\text{a}}=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}$
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