Question

rationalize the denominator in 12 root 2 upon root 3 + root 6​

Answer 1

$\huge{\mathfrak{\blue{\underline{Answer}}}}$

$= \frac{12 \sqrt{2} }{ \sqrt{3} + \sqrt{6} } \\ = \frac{12 \sqrt{2} }{ \sqrt{3} + \sqrt{6} } \times \frac{ \sqrt{3} - \sqrt{6} }{ \sqrt{3} - \sqrt{6} } \\ = \frac{12 \sqrt{6} - 12 \sqrt{12} }{( \sqrt{3}) {}^{2} - ( \sqrt{6} ) {}^{2} } \\ = \frac{12 \sqrt{6} - 12\sqrt{12} }{3 - 6} \\ = \frac{12 \sqrt{6} - 12 \sqrt{12} }{ - 3} \\ = \frac{12( \sqrt{6} - \sqrt{12} ) }{ - 3} \\ = \frac{4( \sqrt{6} - \sqrt{12}) }{ - 1} \\ = 4 \sqrt{6} - 4 \sqrt{12}$

Hence

$\huge{\mathfrak{\blue{\underline{Rationalized}}}}$

Answer 2

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