ABC26GN1441 · Square Roots and Cube Roots

Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: · Marks: · Difficulty:

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}$ is equal to
(a)1
(b)2
(c)$6-\sqrt{35}$
(d)$6+\sqrt{35}$
Answer
Answer (as printed): D
Explanation
$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{(\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5})} \times \frac{(\sqrt{7}+\sqrt{5})}{(\sqrt{7}+\sqrt{5})}=\frac{(\sqrt{7}+\sqrt{5})^{2}}{(7-5)}$ $$=\frac{7+5+2 \sqrt{35}}{2}=\frac{12+2 \sqrt{35}}{2}=6+\sqrt{35} .$$

Explanation as extracted from the printed page; notation may be imperfect.

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