If n is positive, which of the following is equal to

Question

If n is positive, which of the following is equal to  \frac{1}{\sqrt{n+1}-\sqrt{n}}

A. 1

B.  \sqrt{2n+1}

C.  \frac{\sqrt{n+1}}{\sqrt{n}}

D.  \sqrt{n+1}-\sqrt{n}

E.  \sqrt{n+1}+\sqrt{n}

Bunuel · Accepted Answer

If n is positive, which of the following is equal to  \frac{1}{\sqrt{n+1}-\sqrt{n}}

A. 1

B.  \sqrt{2n+1}

C.  \frac{\sqrt{n+1}}{\sqrt{n}}

D.  \sqrt{n+1}-\sqrt{n}

E.  \sqrt{n+1}+\sqrt{n}

This question is dealing with rationalisation of a fraction. Rationalisation is performed to eliminate irrational expression in the denominator. For this particular case we can do this by applying the following rule:  (a-b)(a+b)=a^2-b^2 .

Multiple both numerator and denominator by  \sqrt{n+1}+\sqrt{n} :  \frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}=\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1})^2-(\sqrt{n})^2)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}=\sqrt{n+1}+\sqrt{n} .

Answer: E.

KarishmaB · Answer

When there is an irrational number in the denominator, you rationalize it by multiplying it with its complement i.e. if it is  \sqrt{a} + \sqrt{b}  in the denominator, you will multiply by  \sqrt{a} - \sqrt{b} . This is done to use the algebraic identity (a + b)(a - b) = a^2 - b^2. When a and b are irrational, a^2 and b^2 become rational (given we are dealing with only square roots)

To keep the fraction same, you need to multiply the numerator with the same number as well.

An example will make it clear:

Rationalize

\frac{3}{{\sqrt{2} - 1}}

=  \frac{3}{{\sqrt{2} - 1}} * \frac{\sqrt{2} + 1}{\sqrt{2} + 1}

=  \frac{3*(\sqrt{2} + 1)}{(\sqrt{2})^2 - 1^2}

=  \frac{3*(\sqrt{2} + 1)}{2 - 1}

The denominator has become rational.

Similarly, if the denominator has  \sqrt{a} - \sqrt{b} , you will multiply by  \sqrt{a} + \sqrt{b} .

In this question too, you can substitute n = 1. The given expression becomes  \frac{1}{{\sqrt{2} - 1}} 
Rationalize it and you will get  \sqrt{2} + 1 . Put n = 1 in the options. Only option (E) gives you  \sqrt{2} + 1 .

kalita · Answer

Bunuel - just wanted to clarify an aspect of the roots - the final answer of this problem is E and it is perfectly understood. However, if I want to simplify the  \sqrt{n+1} + \sqrt{n}  even more... theoretically I could "unroot" these expressions, so that I get  2n+1 , however, as the answer B is clearly wrong (and I can see why), I want to but I struggle to understand how to "put the roots back" in the  2n+1  to get an equivalent of  \sqrt{n+1} + \sqrt{n} . Any thoughts on this matter?

Thanks!

Bunuel · Answer

I don't understand what you mean: how can you get  2n+1  from  \sqrt{n+1}+\sqrt{n} ?

kalita · Answer

I meant some people might get  \sqrt{2n+1}  which is the answer B. However, I can see why  \sqrt{n+1}+\sqrt{n}  is NOT equal  \sqrt{2n+1}  even though it might be tempting to simplify it to this form (and pick the wrong answer). But my question is can we simplify  \sqrt{n+1}+\sqrt{n}  further by "squaring" both terms and then "unsquaring" them/the expression back somehow... or what could be an equivalent of  \sqrt{n+1}+\sqrt{n} ?

Bunuel · Answer

\sqrt{n+1}+\sqrt{n}  is the simplest form. If you square it you'll get  (\sqrt{n+1}+\sqrt{n})^2=(\sqrt{n+1})^2+2\sqrt{n+1}*\sqrt{n}+\sqrt{n}^2=2n+1+2\sqrt{(n+1)n} . You cannot take square root from this expression to get anything better than  \sqrt{n+1}+\sqrt{n} .

Hope it's clear.

mbaiseasy · Answer

\frac{1}{\sqrt{n+1}-\sqrt{n}}

\frac{1}{\sqrt{n+1}-\sqrt{n}} * \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}

\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}

\frac{\sqrt{n+1}+\sqrt{n}}{1}

Answer: E

pharm · Answer

I have a quick question on this ..when the initial fraction was rationalized you used:

\sqrt{n+1}+ \sqrt{n} / \sqrt{n+1}+ \sqrt{n}

did you change the sign from negative to positive since the question stated "n" is a positive number. Wouldn't you have to use the same denominator when Rationalizing a fraction?

jlgdr · Answer

Isn't it much easier to just pick n=1 and then look for target in answer choices? Cheers! J

KarishmaB · Answer

Yes, absolutely it is. I would answer this question by plugging in the values but you have to be careful of two things. When pluggin in values in the options, two or more options might seem to satisfy. If this happens, you need to plug in a different number in those two to get the actual correct answer. 
Also, you need to ensure that the value given by option actually does not match the required value before discarding it.
e.g. here if I put n = 1,  \frac{1}{\sqrt{n+1}-\sqrt{n}}  =  \frac{1}{\sqrt{2}-1}

while option (E) gives  \sqrt{n+1}+\sqrt{n}  =  \sqrt{2}+1

You cannot discard option (E) because it doesn't look the same. You must rationalize the value obtained from the expression and then compare it with what you get from option (E). So you must be careful.

Hero8888 · Answer

I put n=3. In that case I got total 3. Only E gives 3 when you put in n=3.

Tip: Don't mess with algebra when you are able to plug-in numbers.

Ritwickhota · Answer

Hi, I had a doubt. Can we solve this particular question via substitution for n. Example: let n=2 and so on?

Harsh9676 · Answer

Yes, But i think algebra is better for this specific question and its often that GMAT never leaves a surd as reminder. Also be careful which number you chose. When you substitute 2 in the question we get (1/\sqrt{3}-\sqrt{2}) and you need to know what's the value for \sqrt{3} and \sqrt{2} >>>>> Suggestion, remember first square roots of first 10 positive numbers . So, the expression becomes = 1/(1.73-1.41) = 3.13 Only choice E gives the closest answer. Hope this helps...

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If n is positive, which of the following is equal to

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If n is positive, which of the following is equal to

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