ABC26GN1440 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: 2010 · Marks: · Difficulty:
$\left[\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}-\sqrt{12}}\right]=$ ?
(a)$\sqrt{3}-\sqrt{2}$
(b)$\sqrt{3}+\sqrt{2}$
(c)$5 \sqrt{3}$
(d)1
Answer
Explanation
Given $\exp .=\frac{3 \sqrt{2}}{(\sqrt{6}-\sqrt{3})} \times \frac{(\sqrt{6}+\sqrt{3})}{(\sqrt{6}+\sqrt{3})}-\frac{4 \sqrt{3}}{(\sqrt{6}-\sqrt{2})}$ $$\begin{aligned} & \times \frac{(\sqrt{6}+\sqrt{2})}{(\sqrt{6}+\sqrt{2})}-\frac{6}{2(\sqrt{2}-\sqrt{3})} \\ = & \frac{3 \sqrt{2}(\sqrt{6}+\sqrt{3})}{(6-3)}-\frac{4 \sqrt{3}(\sqrt{6}+\sqrt{2})}{(6-2)} \\ & \quad+\frac{3}{(\sqrt{3}-\sqrt{2})} \times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})} \\ = & \sqrt{2}(\sqrt{6}+\sqrt{3})-\sqrt{3}(\sqrt{6}+\sqrt{2})+3(\sqrt{3}+\sqrt{2}) \\ = & \sqrt{12}+\sqrt{6}-\sqrt{18}-\sqrt{6}+3 \sqrt{3}+3 \sqrt{2} \\ = & 2 \sqrt{3}-3 \sqrt{2}+3 \sqrt{3}+3 \sqrt{2}=5 \sqrt{3} \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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