Find the value of a in the following :

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Question

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

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

$\frac{6}{3 \sqrt{2}-2 \sqrt{3}}=\frac{6}{3 \sqrt{2}-2 \sqrt{3}} 	imes \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}$

$=\frac{6(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^{2}-(2 \sqrt{3})^{2}}=\frac{6(3 \sqrt{2}+2 \sqrt{3})}{18-12}=\frac{6(3 \sqrt{2}+2 \sqrt{3})}{6}$

$=3 \sqrt{2}+2 \sqrt{3}$

Therefore, $3 \sqrt{2}+2 \sqrt{3}=3 \sqrt{2}-a \sqrt{3}$

or $a=-2$

Find the value of $a$ in the following :

$\frac{6}{3 \sqrt{2}-2 \sqrt{3}}=3 \sqrt{2}-a \sqrt{3}$

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