Question

. Rationalize the denominator and find the value of a and b



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Answer 1

answer in the attachment

Answer 2

$\sf \frac{3 \sqrt{2} }{3 \sqrt{2} - 2 \sqrt{6} } = a + b \sqrt{12}$

$\sf{ \frac{3 \sqrt{2} (3 \sqrt{2} +2 \sqrt{6} )}{(3 \sqrt{2} - 2 \sqrt{6} )(3 \sqrt{2} + 2 \sqrt{6} )} } = a + b \sqrt{12}$

$\sf{ \frac{36 + 6 \sqrt{12} }{ {(3 \sqrt{2} )}^{2} - {(2 \sqrt{6}) }^{2} } } = a + b \sqrt{12}$

$\sf \frac{36 + 6 \sqrt{12} }{36 - 24} = a + b \sqrt{12}$

$\sf \frac{6(6+ \sqrt{12}) }{12} = a + b \sqrt{12}$

$\sf\frac{6 + \sqrt{12} }{2} = a + b \sqrt{12}$

Comparison of both we get,

$\sf \: a \rightarrow 6 \: and \: b \rightarrow \frac{1}{2}$