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Callisto
Group Title
Tutor question #3
Simplify:
\[\sqrt{5\sqrt{21}}\]
(conditions: pen, paper, no calculator, done in 2 minutes)
*
 2 years ago
 2 years ago
Callisto Group Title
Tutor question #3 Simplify: \[\sqrt{5\sqrt{21}}\] (conditions: pen, paper, no calculator, done in 2 minutes) *
 2 years ago
 2 years ago

This Question is Closed

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
what if x = \(\sqrt{5\sqrt{21}}\) then \(x^2 = 5  \sqrt{21}\) \[x^2  5 = \sqrt{21}\] \[(x^2  5)^2 = 21\] oh goodness quadratic o.O
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
actually it is a 4th degree equation
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
\[(x^25)^2=21\]\[x^410x^2+25=21\]\[x^410x^2+4=0\]\[x^410x^2+4=0\]\[x^2=\frac{10\pm \sqrt{(10)^64(4)}}{2}\]\[x^2=5\pm \sqrt{21}\]Sorry... are you sure that it is 'simplified'?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
Oh.that's power 2, typo
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
apparently it is also \[\sqrt{\frac{7}{2}}\sqrt{\frac{3}{2}}\] i remember seeing something like this before but i am not sure i remember how to go from one to the other
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
I need to know how to work that out.....
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
cant seem to get it, i think it was a trick
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
............................ It... was ... a ... question ... asked ... when ... my friend applied for a summer job ...
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
really? doing what??
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
tutor  teaching high school students
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
hmm i wonder what answer they wanted
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
Same here...
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
When I was doing some exercises few days ago, I saw similar questions, but clearer, like this: express \(\sqrt{282\sqrt{147}}\) in the form of \(\sqrt{x}\sqrt{y}\). I can still handle this. But that one, I failed :(
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
multiplying by the conjugate give \(\frac{2}{\sqrt{5+\sqrt{21}}}\) think
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
any examples anywhere?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
You want to show \[\sqrt{5\sqrt{21}} \text{ equals } \frac{\sqrt{7}\sqrt{3}}{2} ?\] was just wondering if you have an example for writing that one thing in that other form
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
oops sqrt(2) on bottom
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
how did you get that identity?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
From experience...
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
lol great experience I want to prove that identity lol
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
First, \((\sqrt{a}  \sqrt{b})^2\) = a + b 2\sqrt{ab} For a>b \[\sqrt{5\sqrt{21}} = \sqrt{52\sqrt{\frac{21}{4}}}\] Now, a+b = 5 => a=5b ab = 21/4 (5b)b = 21/4 4b^2 + 20b  21 =0 b=1.5 or b =3.5 (rejected) a = 5  1.5 = 3.5 So, it is \(\sqrt{\frac{7}{2}}\sqrt{\frac{3}{2}}\) Does that make sense?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
ok i see that identity :)
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
that one is easy to prove
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
maybe because it is an actual identity, right? :p
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
Perfect square is perfect :)
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
Does that make sense? Apart from the latex fail...
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
very interesting i wouldn't have thought of that
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
I'm going to post the link to satellite73's post then. He doesn't even come to check..
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
Great work @Callisto :)
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
And thank you for all your time!!!!
 2 years ago
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