Rationalise the denominator of the following: | vedclass

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(B) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}+\sqrt{2})$.$\

Vedclass · Accepted Answer

$5+2 \sqrt{6}$

Vedclass · Answer

(B) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}+\sqrt{2})$.$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} 	imes \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + b^2 + 2ab$ in the numerator:$= \frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$$= \frac{3 + 2 + 2\sqrt{3} 	imes \sqrt{2}}{3 - 2}$$= \frac{5 + 2\sqrt{6}}{1}$$= 5 + 2\sqrt{6}$

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

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