GMAT Practice Question: If x>0, then \frac{1}{\sqrt{2 x}+\sqrt{x}}=
Question
If x>0, then \frac{1}{\sqrt{2 x}+\sqrt{x}}=
- \frac{1}{\sqrt{3 x}}
- \frac{1}{2 \sqrt{2 x}}
- \frac{1}{x \sqrt{2}}
- \frac{\sqrt{2}-1}{\sqrt{x}}
- \frac{1+\sqrt{2}}{\sqrt{x}}
Topics: roots, simplifications & cancellations
Solution
Step 1
We multiply numerator and denominator by the expression \sqrt{2x}-\sqrt{x} to create a difference of squares in the denominator.
\frac{1}{\sqrt{2x}+\sqrt{x}} \times \frac{\sqrt{2x}-\sqrt{x}}{\sqrt{2x}-\sqrt{x}}
Step 2
We simplify the denominator by applying the difference of squares formula.
\frac{\sqrt{2x}-\sqrt{x}}{2x - x}
Step 3
We factor out \sqrt{x} from the numerator.
\frac{(\sqrt{2}-1)\sqrt{x}}{x}
Step 4
We cancel one factor of \sqrt{x} in the numerator and denominator to obtain the final expression.
\frac{(\sqrt{2}-1)\cancel{\sqrt{x}}}{\cancel{\sqrt{x}}\sqrt{x}} = \frac{\sqrt{2}-1}{\sqrt{x}}
Answer
D
