Question

rationalize the denominator
1 / root 3 + root 6

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Answer 1

$\rm \dfrac{1}{\sqrt{3}+\sqrt{6}}$

We rationalise the denominator by multiplying it's conjugate by itself and numerator.

Here the denominator is $\sqrt{3} + \sqrt{6}$

Conjugate of $\sqrt{3} + \sqrt{6}$ is $\sqrt{3} -\sqrt{6}$

$\rm \therefore \quad \dfrac{1}{\sqrt{3}+\sqrt{6}} \times \dfrac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}$

$\rm \dashrightarrow\dfrac{\sqrt{3}-\sqrt{6}}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})} = \dfrac{\sqrt{3}-\sqrt{6}}{(\sqrt{3})^2 - (\sqrt{6})^2} \qquad ...[since, \ (a+b)(a-b)=a^2-b^2]$

$\rm \dashrightarrow \dfrac{\sqrt{3}-\sqrt{6}}{3- 6}= \dfrac{\sqrt{3}-\sqrt{6}}{-3}$

$Or$

$\rm \dashrightarrow \dfrac{-(\sqrt{3}-\sqrt{6})}{3} = \dfrac{-\sqrt{3} + \sqrt{6}}{3}$

$\dashrightarrow\bf \dfrac{\sqrt{6}-\sqrt{3}}{3}$

Hence, rationalised.

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