ABC26GN1446 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: 2005 · Marks: · Difficulty:
If $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, then $(x+y)$ equals
(a)$2(\sqrt{5}+\sqrt{3})$ $3 \sqrt{2}$ ] $1-\sqrt{5}+\sqrt{2}+\sqrt{10}$
(b)$2 \sqrt{15}$ $\frac{\sqrt{2}}{3}$ ] $1+\sqrt{5}+\sqrt{2}-\sqrt{10}$
(c)8 $\frac{2}{3}$ ] $1+\sqrt{5}-\sqrt{2}+\sqrt{10}$
(d)16 \item[(C.] P.O., 2006) $\frac{1}{3}$ \item[(S.] S.C., 2007) \item[ \item[ \item[ \item[ ] $1-\sqrt{5}-\sqrt{2}+\sqrt{10}$
Answer
Explanation
x + y = + = × 5– 3 5 + 3 ( 5 – 3) ( 5 + 3) ( 5 – 3) ( 5 – 3) + × ( 5 + 3) ( 5 – 3) ( 5 + 3)2 ( 5 – 3)2 = + ( 5)2 – ( 3) 2 ( 5)2 – ( 3)2 ( 5 + 3)2 + ( 5 – 3)2 = 5–3 2 ( 5)2 + ( 3)2 = = 5 + 3 = 8. 2 ( 2)2 (2 + 3) (2 – 3) 2 2 2 – ( 3)2
Explanation as extracted from the printed page; notation may be imperfect.
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