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satellite73
 3 years ago
prove \(\sqrt{5\sqrt{21}}=\frac{\sqrt{7}\sqrt{3}}{\sqrt{2}}\)
satellite73
 3 years ago
prove \(\sqrt{5\sqrt{21}}=\frac{\sqrt{7}\sqrt{3}}{\sqrt{2}}\)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0any help ? i have seen something similar. here is what wofram says http://www.wolframalpha.com/input/?i=sqrt%285sqrt%2821%29%29

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0Aren't you asking the same question in different form?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for two reasons. it might be easier if we know the answer and "simplify" doesn't mean anything in this context

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, but one simple question: in reality, is it possible to do it backward?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this looks for all the world like one of those trig questions where you use one half angle formula and another half angle formula and you get two different looking but equal answers. then it is a pain to show they are the same, but it usually amounts to multiplying repeatedly

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok here is a "well known formula from basic algebra" \[\sqrt{a+b\sqrt{c}}=\sqrt{\frac{1}{2}(a+\sqrt{q})}\sqrt{\frac{1}{2}(a\sqrt{q})}\] if \(q=a^2b^2c\) is a perfect square

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i must have missed that day

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0guess i have my homework cut out for me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0typo actually it should be \[\sqrt{a\pm b\sqrt{c}}=\sqrt{\frac{1}{2}(a+\sqrt{q})}\pm\sqrt{\frac{1}{2}(a\sqrt{q})}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0given this formula we are done, since \(5^221=4\) a perfect square, so we get the answer immediately. how we know this i am not sure

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, I don't want to do so. but.. Please check this http://openstudy.com/study#/updates/4fda8ca0e4b0f2662fd103f2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok lets try this start with \[\sqrt{5\sqrt{21}}\] and write the radicand as a perfect square. so \[\sqrt{5\sqrt{21}}=xy=\sqrt{(xy)^2}=\sqrt{x^2+y^22xy}\] now we can put \(x^2+y^2=5, 2xy=\sqrt{21}, y=\frac{\sqrt{21}}{2x}\) and so \(x^2+(\frac{\sqrt{21}}{2x})^2=5\) and so \[4x^420x^2+21=0\] solving gives \(x=\sqrt{\frac{7}{2}}\) i think

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i will have to look at this more the first "well known formula" comes directly from this paper, but no explanation, so not much help except in getting the answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the second method i tried to mimic what i read here http://gauravtiwari.org/2011/03/16/aproblemonordinarynestedradicals/ but the line "which on simplification yields..." was not so obvious to me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, take a look at 4.5.2 here and the following theorem. this i think give it all

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0learn something new every day!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0today for example i learned where to get a really great hamburger

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your friend must have run in to someone who was either nuts ( a nutty math teacher, imagine) or just learned this and wanted to show off. i cannot imagine this in a basic algebra class

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0*Disappointed* You're not learning HOW to get a great hamburger but you learnt WHERE to get it :( Thanks for all the notes and the 'well unknown formula' which I haven't heard of. Hmm.. Actually, I can solve it :

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0great. let me know how it goes. if they had had the internet when i was in school i would have 4 degrees by now

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0This might be good enough. Look at \(\sqrt{5\sqrt{21}}\). If you square it, you get \(5\sqrt{21}\). We want this to be a perfect square of some binomial with square roots. This means, that we should want a square multiplied by two. Hence, multiply this by two. We get \[102\sqrt{21}\]If we want this to be a perfect square of some expression that looks like \(\sqrt x\pm\sqrt y\), we want to find an \(x, y\) such that \(x+y=10\) and \(xy=21\). Solving this, we get that \[102\sqrt{21}=(\sqrt7\sqrt3)^2\]Now we work backward to get what we had at the start. Divide by 2, and then take the square root gives us that \[\sqrt{5\sqrt{21}}=\frac{\sqrt7\sqrt3}{\sqrt2}\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Typo in that first paragraph. I said "This means, that we should want a square multiplied by two." It should read "This means, that we should want a square root multiplied by two." (talking about the square root in \(5\sqrt{21}\) of course)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge I solved the question using this way. But no one here looks at it except myininaya

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this looks much snappier

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0The two methods are very slightly different. Same trick, but slightly different process.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Namely, I took things out of the square root, and you left them in. So I guess they really are just the same thing.
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