If k > 1, which of the following must be equal to...

[swerve]

If k > 1, which of the following must be equal to
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}?$$

$$A.\ 2$$
$$B.\ 2\sqrt{k}$$
$$C.\ 2\sqrt{k+1}+\sqrt{k-1}$$
$$D.\ \frac{\sqrt{k+1}}{\sqrt{k-1}}$$
$$E.\ \sqrt{k+1}-\sqrt{k-1}$$

The OA is E.

I know that I can solve this PS question in an easy way like,
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\cdot\frac{\sqrt{k+1}-\sqrt{k-1}}{\sqrt{k+1}-\sqrt{k-1}}=\frac{2\sqrt{k+1}-\sqrt{k-1}}{2}=\sqrt{k+1}-\sqrt{k-1}$$
Please, can any expert explain this PS question for me? I would like to know another way to solve it. I need your help. Thanks.

[GMATGuruNY]

If k > 1, which of the following must be equal to
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}?$$

$$A.\ 2$$
$$B.\ 2\sqrt{k}$$
$$C.\ 2\sqrt{k+1}+\sqrt{k-1}$$
$$D.\ \frac{\sqrt{k+1}}{\sqrt{k-1}}$$
$$E.\ \sqrt{k+1}-\sqrt{k-1}$$

√2 ≈ 1.4.
√3 ≈ 1.7.

Let k=2.
Plugging k=2 into the given expression, we get:
2/(√3 + √1) ≈ 2/(1.7 + 1) = 2/(2.7) = 20/27 ≈ 20/28 = 5/7.

Now plug k=2 into the answer choices to see which yields a value close to 5/7.
A quick scan reveals the following:
The results from A, B, C and D will all be greater than 1.

The correct answer is E.

If we plug k=2 into E, we get:
√(k+1) - √(k-1) ≈ √3 - √1 = 1.7 - 1 = 0.7 = 7/10.

[[email protected]]

Hi swerve,

The restriction that K > 1 actually has no impact on this question. If you choose K=1, then you'll end up 2/√2 = √2.

There's only one answer that equals √2 when you TEST K=1...

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

If k > 1, which of the following must be equal to
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}?$$

$$A.\ 2$$
$$B.\ 2\sqrt{k}$$
$$C.\ 2\sqrt{k+1}+\sqrt{k-1}$$
$$D.\ \frac{\sqrt{k+1}}{\sqrt{k-1}}$$
$$E.\ \sqrt{k+1}-\sqrt{k-1}$$

Recall that the conjugate of an expression replaces the "plus" sign between the terms with a "minus" sign. Thus, the conjugate of √(k + 1) + √(k - 1) is √(k + 1) - √(k - 1). Multiplying both the numerator and the denominator of the given expression by the conjugate of the denominator allows us to rationalize the denominator (i.e., rid the denominator of any radical signs). Thus, we have:
2[√(k + 1) - √(k - 1)]/{[√(k + 1) + √(k - 1)][√(k + 1) - √(k - 1)]}

2[√(k + 1) - √(k - 1)]/[(k + 1) - (k - 1)]

2[√(k + 1) - √(k - 1)]/2

√(k + 1) - √(k - 1)

Answer: E