Rationalise the denominator of the following: | vedclass

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Question

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{2}-\sqrt{3})$:$\fra

Vedclass · Accepted Answer

$3 \sqrt{2}-2 \sqrt{3}$

Vedclass · Answer

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{2}-\sqrt{3})$:$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} 	imes \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:$= \frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2}$$= \frac{\sqrt{12} - \sqrt{18}}{2 - 3}$$= \frac{\sqrt{12} - \sqrt{18}}{-1}$$= \sqrt{18} - \sqrt{12}$$= \sqrt{9 	imes 2} - \sqrt{4 	imes 3}$$= 3\sqrt{2} - 2\sqrt{3}$

Rationalise the denominator of the following:
$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

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Rationalise the denominator of the following:$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$