anonymous
  • anonymous
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))))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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phi
  • phi
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
anonymous
  • anonymous
The question is to rationalize the first one Choice A
phi
  • phi
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?

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More answers

anonymous
  • anonymous
on the top you would get 2 and the bottom would be 6+4sqrt(2)
phi
  • phi
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)
anonymous
  • anonymous
top would be \[(\sqrt{2-\sqrt{2}})*(\sqrt{2+\sqrt{2}}) and the bottom would be 2 sqrt(2)
anonymous
  • anonymous
bottom\[2+\sqrt{2}\]
phi
  • phi
For the top we can use sqrt(A)*sqrt(B)= sqrt(A*B) . That means we get \[\frac{\sqrt{(2-\sqrt{2})(2+\sqrt{2})}}{2+\sqrt{2}}\]
anonymous
  • anonymous
right
phi
  • phi
Now for the top, we use (a+b)(a-b)= a^2-b^2
phi
  • phi
so we get (with a square root still out there) \[\frac{\sqrt{2}}{2+\sqrt{2}}\]
phi
  • phi
Agree?
anonymous
  • anonymous
yes..im following
phi
  • phi
now multiply top and bottom by 2-sqrt(2) to get rid of the radical in the denominator
phi
  • phi
\[\frac{\sqrt{2}}{2+\sqrt{2}}\cdot \frac{2-\sqrt{2}}{2-\sqrt{2}}= \frac{2\sqrt{2}-2}{4-2}\]
anonymous
  • anonymous
\[2\sqrt{2}-2/2\]
phi
  • phi
That simplifies to sqrt(2)-1 now we remember the outer square root, to get the final (ugly) answer: \[ \sqrt{\sqrt{2}-1} \]
anonymous
  • anonymous
Thank you so much. thats crazy...thank you again
phi
  • phi
It is what it is.

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