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If n is positive, which of the following is equal to [#permalink]
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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. \(\sqrt{n+1}+\sqrt{n}\)
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Originally posted by kook44 on 28 Jun 2006, 06:36.
Last edited by Bunuel on 16 Apr 2012, 02:05, edited 1 time in total.
Edited the question and added the OA
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If n is positive, which of the following is equal to [#permalink]
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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.
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Re: fraction [#permalink]
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Quote: 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. 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!
Originally posted by kalita on 17 Nov 2012, 05:42.
Last edited by kalita on 18 Nov 2012, 03:47, edited 2 times in total.
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Re: fraction [#permalink]
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17 Nov 2012, 05:51
ikokurin wrote: Bu nuel wrote: 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. 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 "unsquare" these expressions, so that I get 2n+1, however, as the answer B is clearly wrong (and I can see why), I struggle to understand how to "square back" the 2n+1 to get an equivalent of SQRT(n+1) + SQRT(n). Can you help me out or share your thoughts on the matter? Thanks! I don't understand what you mean: how can you get \(2n+1\) from \(\sqrt{n+1}+\sqrt{n}\)?
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Re: fraction [#permalink]
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Updated on: 18 Nov 2012, 03:39
Quote: I don't understand what you mean: how can you get \(2n+1\) from \(\sqrt{n+1}+\sqrt{n}\)? 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 \(\sqrt{n+1}+\sqrt{n}\)?
Originally posted by kalita on 17 Nov 2012, 06:06.
Last edited by kalita on 18 Nov 2012, 03:39, edited 1 time in total.
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Re: fraction [#permalink]
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17 Nov 2012, 06:19
ikokurin wrote: 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 "squarerooting" them again somehow... or what could be an equivalent of SQRT(n+1) + SQRT(n)? \(\sqrt{n+1}+\sqrt{n}\) 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.
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Re: fraction [#permalink]
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Updated on: 18 Nov 2012, 03:32
Quote: \(\sqrt{n+1}+\sqrt{n}\) 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. I see. What you are saying is clear but your answer does not exactly address what I am after. I can see that \((\sqrt{n+1}+\sqrt{n})^2\) only complicates it further. Sorry to be pertinacious on this - if we do \((\sqrt{n+1})^2+(\sqrt{n})^2\) => we will get \(n+1 + n = 2n + 1\) => can we "undo" the expression \(2n + 1\) somehow to get the equivalent of \(\sqrt{n+1}+\sqrt{n}\)? I promise this is the last one:) P.S. Also, please let me know if it would be better to send a PM on related "clarifying" questions...
Originally posted by kalita on 17 Nov 2012, 06:37.
Last edited by kalita on 18 Nov 2012, 03:32, edited 1 time in total.
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Re: fraction [#permalink]
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17 Nov 2012, 06:45
ikokurin wrote: \(\sqrt{n+1}+\sqrt{n}\) 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. I see. What you are saying is clear but your answer does not exactly address what I am after. Sorry to be pertinacious on this but I was wondering if we can do (SQRT(n+1))^2 + (SQRT(n))^2 => we will get n+1 + n = 2n + 1 => can we "undo" the expression 2n + 1 somehow to get the equivalent of SQRT(n+1) + SQRT(n)? I promise this is the last one:) Also, please let me know if it would be better to send a PM on related "clarifying" questions...[/quote] The answer is no, these expressions are not equal. P.S. Please use formatting, check here: rules-for-posting-please-read-this-before-posting-133935.html#p1096628
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Re: fraction [#permalink]
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17 Nov 2012, 18:44
Quote: The answer is no, these expressions are not equal. P.S. Please use formatting, check here: rules-for-posting-please-read-this-before-posting-133935.html#p1096628I understand they are not equal, thanks for help. So I take away there is no way to go from \(2n+1\) (obtained after squaring both terms (\((\sqrt{n+1})^2 + (\sqrt{n})^2\)) into something else that could be an equivalent of\(\sqrt{n+1} + \sqrt{n}\). As mentioned above, for those having issues with exponents/roots, it is possible to make a mistake of simplifying \((\sqrt{n+1})^2 + (\sqrt{n})^2\) into \(\sqrt{2n+1}\) (which is incorrect); nevertheless I wanted to see if there was a way to do something about \(2n+1\) to make it equal to \(\sqrt{n+1} + \sqrt{n}\). For some reason, having inner desire to combine those \(n\) terms to make it all look nicer, it bugs me that leaving the answer as \(\sqrt{n+1} + \sqrt{n}\) is all we can do about this equation; especially after I saw some tricks/solutions relating to the tricky exponent problems and how one can do "wonders" with squaring and unsquaring things  I was thinking about simplifying this thing into something like, obviously grossly exaggerated, \(^4\sqrt{2n+1}\) or \(\sqrt{2n}+\sqrt{1}\), etc., by "squarerooting" \(2n+1\) back somehow. But again I know the previous examples are plain wrong, just giving an example of what one can go through working through possibilities. Anyhow, enough of this rumble, let me know if you have anything to add...and thanks much for patience. Regards,
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Re: If n is positive, which of the following is equal to [#permalink]
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11 Dec 2012, 04:33
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}\) \(\frac{1}{\sqrt{n+1}-\sqrt{n}} * \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\) \(\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\) \(\frac{\sqrt{n+1}+\sqrt{n}}{1}\) Answer: E
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Re: If n is positive, which of the following is equal to [#permalink]
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23 Jan 2013, 08:27
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?
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Re: If n is positive, which of the following is equal to [#permalink]
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pharm wrote: 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? 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 \(\sqrt{2} + 1\).
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Re: If n is positive, which of the following is equal to [#permalink]
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28 Jan 2013, 08:51
Thanks Karishma that cleared things up
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Re: If n is positive, which of the following is equal to [#permalink]
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22 Nov 2013, 07:12
kook44 wrote: 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. \(\sqrt{n+1}+\sqrt{n}\) Isn't it much easier to just pick n=1 and then look for target in answer choices? Cheers! J
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Re: If n is positive, which of the following is equal to [#permalink]
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22 Nov 2013, 08:01
jlgdr wrote: kook44 wrote: 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. \(\sqrt{n+1}+\sqrt{n}\) Isn't it much easier to just pick n=1 and then look for target in answer choices? Cheers! J What if more than one answer choice gives you same value? first, we have to try original expression with 1 and try each of the choices with 1. If we are lucky we have only one choice matching. but what if there are 2 or even 3 answer choices? we would then have to pick another number. Personally I feel solving it is faster in this case. Sometimes number picking works faster. knowing when to use number picking is the difficult part.
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Re: If n is positive, which of the following is equal to [#permalink]
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22 Nov 2013, 08:29
Ya I guess your right after solving the way Bunuel did it took less than 20 secs
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Re: If n is positive, which of the following is equal to [#permalink]
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24 Nov 2013, 21:05
jlgdr wrote: kook44 wrote: 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. \(\sqrt{n+1}+\sqrt{n}\) Isn't it much easier to just pick n=1 and then look for target in answer choices? Cheers! J 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.
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Re: If n is positive, which of the following is equal to [#permalink]
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30 Jan 2018, 01:47
Bunuel wrote: 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. Hi! I do not understand, how you came up with the last part of the solution where you just simplify and take the expression from the denominator away. thank you!
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Re: If n is positive, which of the following is equal to [#permalink]
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30 Jan 2018, 02:34
Jannnn04 wrote: Bunuel wrote: 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. Hi! I do not understand, how you came up with the last part of the solution where you just simplify and take the expression from the denominator away. thank you! The denominator is n+1-n, which is 1: n + 1 - n= 1.
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Senior Manager
Joined: 09 Mar 2016
Posts: 448
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Re: If n is positive, which of the following is equal to [#permalink]
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25 Mar 2018, 06:15
kook44 wrote: 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 link is great source about rationalizing denominator  http://www.wtamu.edu/academic/anns/mps/ ... nalize.htm
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Re: If n is positive, which of the following is equal to
[#permalink]
25 Mar 2018, 06:15
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