ABC26GN1438 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: 2005 · Marks: · Difficulty:
What is the value of $\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{2}{3+\sqrt{5}}-\frac{3}{3+\sqrt{3}}$ ?
(a)$-\frac{1}{2}$
(b)0
(c)$\frac{1}{2}$
(d)1
Answer
Explanation
Given $\exp .=\frac{1}{(\sqrt{5}+\sqrt{3})} \times \frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})}+\frac{2}{(3+\sqrt{5})}$ $$\begin{aligned} & \times \frac{(3-\sqrt{5})}{(3-\sqrt{5})}-\frac{3}{(3+\sqrt{3})} \times \frac{(3-\sqrt{3})}{(3-\sqrt{3})} \\ = & \frac{(\sqrt{5}-\sqrt{3})}{(5-3)}+\frac{2(3-\sqrt{5})}{(9-5)}-\frac{3(3-\sqrt{3})}{(9-3)} \\ = & \frac{\sqrt{5}-\sqrt{3}}{2}+\frac{2(3-\sqrt{5})}{4}-\frac{3(3-\sqrt{3})}{6} \\ = & \frac{6(\sqrt{5}-\sqrt{3})+6(3-\sqrt{5})-6(3-\sqrt{3})}{12}=0 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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