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 $${{\text{x}}^2} - 1$$

Correct Answer: $${{\text{x}}^2} - 1$$

$$\eqalign{ & \Leftrightarrow \sqrt {x + \frac{x}{y}} = x\sqrt {\frac{x}{y}} \cr & \Leftrightarrow x + \frac{x}{y} = {x^2}.\frac{x}{y} \cr & \Leftrightarrow \frac{{xy + x}}{y} = \frac{{{x^3}}}{y} \cr & \Leftrightarrow xy + x = {x^3} \cr & \Leftrightarrow y + 1 = {x^2} \cr & \Leftrightarrow y = {x^2} - 1 \cr} $$
Bissoy MCQ
#Arithmetic Ability #Square Root And Cube Root

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