Question

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

Answer 1

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

Answer 2

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

Answer 3

yes

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

i must have missed that day

Answer 9

guess i have my homework cut out for me

Answer 10

typo actually it should be
\[\sqrt{a\pm b\sqrt{c}}=\sqrt{\frac{1}{2}(a+\sqrt{q})}\pm\sqrt{\frac{1}{2}(a-\sqrt{q})}\]

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

learn something new every day!

Answer 18

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

Answer 19

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

Answer 20

*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 :|

Answer 21

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

Answer 22

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 \[\sqrt{5-\sqrt{21}}=\frac{\sqrt7-\sqrt3}{\sqrt2}\]

Answer 23

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)

Answer 24

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

Answer 25

this looks much snappier

Answer 26

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

Answer 27

Namely, I took things out of the square root, and you left them in. So I guess they really are just the same thing.