If $$\sqrt {x + \frac{x}{y}} = x\sqrt {\frac{x}{y}} {\text{,}}$$ where x and y are positive real numbers, then y is equal to ?
A. $${\text{x}} + 1$$
B. $${\text{x}} - 1$$
C. $${{\text{x}}^2} + 1$$
D. $${{\text{x}}^2} - 1$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \Leftrightarrow \sqrt {x + \frac{x}{y}} = x\sqrt {\frac{x}{y}} \cr & \Leftrightarrow x + \frac{x}{y} = {x^2}.\frac{x}{y} \cr & \Leftrightarrow \frac{{xy + x}}{y} = \frac{{{x^3}}}{y} \cr & \Leftrightarrow xy + x = {x^3} \cr & \Leftrightarrow y + 1 = {x^2} \cr & \Leftrightarrow y = {x^2} - 1 \cr} $$Related Questions on Square Root and Cube Root
The least perfect square, which is divisible by each of 21, 36 and 66 is:
A. 213444
B. 214344
C. 214434
D. 231444
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