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

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!

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 \(\sqrt{n+1}+\sqrt{n}\)?

\(\sqrt{n+1}+\sqrt{n}\) 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...

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

I 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,

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

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


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

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




Answer: E

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?

Thanks Karishma that cleared things up

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.

Ya I guess your right after solving the way Bunuel did it took less than 20 secs

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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.

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.

this link is great source about rationalizing denominator https://www.wtamu.edu/academic/anns/mps/ ... nalize.htm