Question

Rationalise the denominator of
$\frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3 } }$

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Answer 1

$\frac{6 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \times \frac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} }$

$\frac{(6 \sqrt{3})( \sqrt{6} + \sqrt{2}) }{( \sqrt{6} - \sqrt{2})( \sqrt{6} + \sqrt{2}) }$

$\frac{6 \sqrt{18} + 6 \sqrt{6} }{6 - 2}$

$\frac{6( \sqrt{9 \times 2 } + \sqrt{3 \times 2 )} }{4}$

$\frac{6 \times 3 \times 2( \sqrt{3}) }{4}$

$\frac{36 \sqrt{3} }{4}$

$9 \sqrt{3}$

Answer 2

Answer:

$6$

Step-by-step explanation:

$\frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \times \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } \\ = \frac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} ) }{ (\sqrt{6} - \sqrt{3} ) ^{2} } \\ = \frac{ \sqrt{9 \times 2}( \sqrt{6} + \sqrt{3}) }{( \sqrt{6} - \sqrt{3} ) ^{2} } \\ = \frac{ \sqrt{18} \times \sqrt{6} \times \sqrt{3} }{6 - 3} \\ = \frac{ \sqrt{18} \times \sqrt{18} }{3} \\ = \frac{18}{3} \\ = 6$