Question

Help with algebra, radicals:


http://imageshack.us/photo/my-images/401/radi1.png/


Does anyone has any idea in order to find x/y?
The calculator didn't help.
Thanks.

Answer 1

i cant open the link, can u olz post the question? :)

Answer 2

I'll Try to post it.

Answer 3

Given
\[x=\sqrt{26-\sqrt{7}}\] and \[y=\sqrt{3-\sqrt{7}}\]
The division x/y is equivalent to the expression:
\[a+\sqrt{b}\]
Where a and b are positive integers, then
\[a^2-b\]
is equal to:

a)9 b)15 c)29
d)2 e)18

Answer 4

*x =\[\sqrt{26-2\sqrt{7}}\]

Answer 5

id probably say rationalize denum of inside in\[\frac{x}{y}=\sqrt{\frac{26-2\sqrt{7}}{3-\sqrt{7}}}\]have u tried it? :)

Answer 6

Yes, and i get

\[\frac{ x }{ y }=\sqrt{10\sqrt{7}+32}\]

But what should i do with this, for find a and b?

Answer 7

very good till here, see it will be something like\[\frac{ x }{ y }=\sqrt{10\sqrt{7}+32}=\sqrt{(a+\sqrt{b})^2}\]and u want to find a and b, so remember this simple identity\[(a+\sqrt{b})^2=a^2+2a\sqrt{b}+b\]

Answer 8

what do u think? :)

Answer 9

Well, then it is supossed that
\[a^2+b= 32\]
and
\[2a \sqrt{b} = 10\sqrt{7}\]
Can b be different of 7?

Answer 10

there are many infinitely answers for the equation\[\sqrt{10\sqrt{7}+32}=\sqrt{(a+\sqrt{b})^2}\]and yes b can be different from 7, but the one u can see it clearly is b=7 :)

Answer 11

Well if b = 7, then a = 5 and a^2-b = 18, and that matches with alternative e.
But if that ecuation has infinety solutions, couldn't the other alternatives be also correct?

Answer 12

usually thats what this types of questions wants from u (b=7 stuff), emm...i dont know maybe. we should check it somehow.

Answer 13

That´s logical, anyway, i think that the system:
\[a^2+b=32\]
\[2a \sqrt{b}=2\sqrt{7}\]

Is only truth if, a = 5 and b = 7

Answer 14

ok, i'll back to this later :) but we accept a=5 and b=7 by now.