Question

rationalize the denominator root 6 by root 3 minus root 2​

Answer 1

Answer :

3√6+4√3

Step-by-step explanation:

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