prove \(\sqrt{5-\sqrt{21}}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)

- satellite73

prove \(\sqrt{5-\sqrt{21}}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)

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- chestercat

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- anonymous

any help ? i have seen something similar. here is what wofram says
http://www.wolframalpha.com/input/?i=sqrt%285-sqrt%2821%29%29

- Callisto

Aren't you asking the same question in different form?

- anonymous

yes

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- anonymous

for two reasons.
it might be easier if we know the answer and
"simplify" doesn't mean anything in this context

- Callisto

Sorry, but one simple question:
in reality, is it possible to do it backward?

- anonymous

this 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

ok 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 \(q=a^2-b^2c\) is a perfect square

- anonymous

i must have missed that day

- anonymous

guess i have my homework cut out for me

- anonymous

typo 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

given this formula we are done, since \(5^2-21=4\) a perfect square, so we get the answer immediately.
how we know this i am not sure

- Callisto

Sorry, I don't want to do so. but..
Please check this
http://openstudy.com/study#/updates/4fda8ca0e4b0f2662fd103f2

- anonymous

ok lets try this
start with
\[\sqrt{5-\sqrt{21}}\]
and write the radicand as a perfect square. so
\[\sqrt{5-\sqrt{21}}=x-y=\sqrt{(x-y)^2}=\sqrt{x^2+y^2-2xy}\]
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
\[4x^4-20x^2+21=0\] solving gives \(x=\sqrt{\frac{7}{2}}\) i think

- anonymous

i 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

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- anonymous

the second method i tried to mimic what i read here
http://gauravtiwari.org/2011/03/16/a-problem-on-ordinary-nested-radicals/
but the line "which on simplification yields..." was not so obvious to me

- anonymous

ok, take a look at 4.5.2 here and the following theorem. this i think give it all

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- anonymous

learn something new every day!

- anonymous

today for example i learned where to get a really great hamburger

- anonymous

your 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

*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

great. 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

This 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 \[10-2\sqrt{21}\]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 \[10-2\sqrt{21}=(\sqrt7-\sqrt3)^2\]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

Typo 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

@KingGeorge I solved the question using this way. But no one here looks at it except myininaya

- anonymous

this looks much snappier

- KingGeorge

The two methods are very slightly different. Same trick, but slightly different process.

- KingGeorge

Namely, 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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