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Powers and Roots Practice Test 5

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Question : 4 of 31

Marks: +1, -0

\left[\frac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}-\sqrt{12}}\right]

is equal to:

1
$5\sqrt{3}$
$\sqrt{3}+\sqrt{2}$
$\sqrt{3}-\sqrt{2}$

Solution:

Concept:

Rationalise each denominator using its conjugate, then simplify.

Explanation:

First term:

\frac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}

Multiply by

\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}

.

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

.

Second term:

\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}

Multiply by

\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}

.

=\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{6-2}=\sqrt{3}(\sqrt{6}+\sqrt{2})=3\sqrt{2}+\sqrt{6}

.

Third term:

\frac{6}{\sqrt{8}-\sqrt{12}}

Since

\sqrt{8}=2\sqrt{2}

and

\sqrt{12}=2\sqrt{3}

,

=\frac{6}{2(\sqrt{2}-\sqrt{3})}=\frac{3}{\sqrt{2}-\sqrt{3}}

.

Rationalise:

\frac{3(\sqrt{2}+\sqrt{3})}{2-3}=-3(\sqrt{2}+\sqrt{3})

.

Now combine:

(2\sqrt{3}+\sqrt{6})-(3\sqrt{2}+\sqrt{6})-[-3(\sqrt{2}+\sqrt{3})]

=2\sqrt{3}+\sqrt{6}-3\sqrt{2}-\sqrt{6}+3\sqrt{2}+3\sqrt{3}

=5\sqrt{3}

.

Answer:

5\sqrt{3}

(Option B).

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