Question

I am having a hard time trying to rationalize the following problem please help.

The square root of the square root of 2- the square root of 2 divided by the square root of 2+ the square root of 2. (if anyone knows how i can type it in word and transfer it let me know but roughly this what it looks like: √((√(2-√(2)))/(√(2+√(2))))

Answer 1

is the question rationalize A:
\[\sqrt{\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}}\]

or B:
\[\frac{\sqrt{\sqrt{2-\sqrt{2}}}}{\sqrt{2+\sqrt{2}}} \]

or C:
neither

Answer 2

The question is to rationalize the first one Choice A

Answer 3

First step is ignore the outer square root (for the moment. Remember it is there, but concentrate on the stuff inside)
multiply top and bottom by sqrt(2+sqrt(2))
what do you get?

Answer 4

on the top you would get 2 and the bottom would be 6+4sqrt(2)

Answer 5

OK, I am ignoring the outer square root (but it is still there hovering over us)
\[\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\cdot \frac{\sqrt{2+\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\]
think of the bottom as sqrt(A)*sqrt(A). It is A (in this case a complicated A)

Answer 6

top would be \[(\sqrt{2-\sqrt{2}})*(\sqrt{2+\sqrt{2}}) and the bottom would be 2
sqrt(2)

Answer 7

bottom\[2+\sqrt{2}\]

Answer 8

For the top we can use sqrt(A)*sqrt(B)= sqrt(A*B) . That means we get
\[\frac{\sqrt{(2-\sqrt{2})(2+\sqrt{2})}}{2+\sqrt{2}}\]

Answer 9

right

Answer 10

Now for the top, we use (a+b)(a-b)= a^2-b^2

Answer 11

so we get (with a square root still out there)
\[\frac{\sqrt{2}}{2+\sqrt{2}}\]

Answer 12

Agree?

Answer 13

yes..im following

Answer 14

now multiply top and bottom by 2-sqrt(2) to get rid of the radical in the denominator

Answer 15

\[\frac{\sqrt{2}}{2+\sqrt{2}}\cdot \frac{2-\sqrt{2}}{2-\sqrt{2}}= \frac{2\sqrt{2}-2}{4-2}\]

Answer 16

\[2\sqrt{2}-2/2\]

Answer 17

That simplifies to sqrt(2)-1

now we remember the outer square root, to get the final (ugly) answer:
\[ \sqrt{\sqrt{2}-1} \]

Answer 18

Thank you so much. thats crazy...thank you again

Answer 19

It is what it is.