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Powers and Roots Practice Test 4

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Question : 9 of 40

Marks: +1, -0

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

-\frac{6}{\sqrt{8}+\sqrt{12}}=?

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

Solution:

Concept:

Rationalize each denominator separately, then simplify the combined expression.

Explanation:

For the first term:

T_1=\frac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}\times \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}=2\sqrt{3}+\sqrt{6}

For the second term:

T_2=\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}\times \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=3\sqrt{2}+\sqrt{6}

For the third term:

T_3=\frac{6}{\sqrt{8}+\sqrt{12}}=\frac{6}{2(\sqrt{2}+\sqrt{3})}=\frac{3}{\sqrt{2}+\sqrt{3}}=3(\sqrt{3}-\sqrt{2})

Now substitute:

T_1-T_2-T_3=(2\sqrt{3}+\sqrt{6})-(3\sqrt{2}+\sqrt{6})-3(\sqrt{3}-\sqrt{2})

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

=-\sqrt{3}

Answer:

The correct answer is

-\sqrt{3}

, option B.

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