# To rationalize the denominator of the expression \frac{3}{2-\sqrt{3}}

To rationalize the denominator of the expression $$\displaystyle{\frac{{{3}}}{{{2}-\sqrt{{{3}}}}}}$$

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Faiza Fuller
Step 1
Multiplying the numerator and the denominator with the conjugate,
$$\displaystyle{\frac{{{3}}}{{{2}-\sqrt{{{3}}}}}}={\frac{{{3}}}{{{2}-\sqrt{{{3}}}}}}\times{\frac{{{2}+\sqrt{{{3}}}}}{{{2}+\sqrt{{{3}}}}}}$$
$$\displaystyle{\frac{{{3}}}{{{2}-\sqrt{{{3}}}}}}={\frac{{{\left({3}\right)}{\left({2}+\sqrt{{{3}}}\right)}}}{{{\left({2}-\sqrt{{{3}}}\right)}{\left({2}+\sqrt{{{3}}}\right)}}}}$$
$$\displaystyle{\frac{{{3}}}{{{2}-\sqrt{{{3}}}}}}={\frac{{{6}+{3}\sqrt{{{3}}}}}{{{\left({2}\right)}^{{{2}}}-{\left(\sqrt{{{3}}}\right)}^{{{2}}}}}}\ {i}.{e}.\ {\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{{2}}}-{b}^{{{2}}}$$
$$\displaystyle{\frac{{{3}}}{{{2}-\sqrt{{{3}}}}}}={\frac{{{6}+{3}\sqrt{{{3}}}}}{{{4}-{3}}}}$$ (Since square and square root are opposite operations, they cancel out)
$$\displaystyle{\frac{{{3}}}{{{2}-\sqrt{{{3}}}}}}={6}+{3}\sqrt{{{3}}}$$