Question

If $0<y<x,$ which statements must be true?

• A. a) $\sqrt{x}-\sqrt{y}=\sqrt{x-y}$
• B. b) $x\sqrt{y}=y\sqrt{x}$
• C. c) $\sqrt{x}+\sqrt{x}=\sqrt{2x}$
• D. d) $\sqrt{xy}=\sqrt{x}\sqrt{y}$.

Answer 1

The correct option is D

d) $\sqrt{xy}=\sqrt{x}\sqrt{y}$.

Explanation for the correct option

Option (d): $\sqrt{xy}=\sqrt{x}\sqrt{y}$

$$\begin{gathered} L.H.S=\sqrt{xy} \\ ={\left(xy\right)}^{\frac{1}{2}} \\ =x^{\frac{1}{2}}.y^{\frac{1}{2}}[\because {(a.b)}^m=a^m.b^m] \\ =\sqrt{x}.\sqrt{y} \\ =R.H.S \end{gathered}$$

Thus, $\sqrt{xy}=\sqrt{x}\sqrt{y}$ is a true statement.

Explanation for the incorrect options

Option (a): $\sqrt{x}-\sqrt{y}=\sqrt{x-y}$

Here, $\sqrt{x}-\sqrt{y}=\sqrt{x-y}$ is a false statement because to add or subtract under root, bases must be same.

Example:

$$\begin{gathered} \sqrt{3}-\sqrt{2}\ne \sqrt{3-2} \\ 1.732....-1.414...\ne 1 \\ 0.317....\ne 1 \end{gathered}$$

Therefore, $\sqrt{x}-\sqrt{y}=\sqrt{x-y}$ is false.

Option (b): $x\sqrt{y}=y\sqrt{x}$

Example:

$$\begin{gathered} L.H.S:4\sqrt{9}=4\times 3 \\ =12 \\ R.H.S:9\sqrt{4}=9\times 2 \\ =18 \\ L.H.S\ne R.H.S \end{gathered}$$

Therefore, $x\sqrt{y}=y\sqrt{x}$ is false.

Option (c): $\sqrt{x}+\sqrt{x}=\sqrt{2x}$

$$\begin{gathered} L.H.S=\sqrt{x}+\sqrt{x} \\ =\sqrt{x}(1+1) \\ =2\sqrt{x} \\ \ne R.H.S \end{gathered}$$

Thus, $\sqrt{x}+\sqrt{x}=\sqrt{2x}$ is a false statement.

Hence, option $\left(d\right)$ is the correct answer.