How to find the square root of a square root?

AI Thread Summary
To simplify √(53 - 8√15), the discussion centers on the method of expressing numbers as √a - √b. The initial attempts involved squaring both sides of an equation, but confusion arose regarding the next steps. After some back and forth, the solution was clarified, leading to the final answer of 4√3 - √5. The exchange highlights the importance of understanding the squaring process in simplifying square roots. Overall, the problem was successfully resolved through collaborative discussion.
Cheesycheese213
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Homework Statement


Simplify √(53 - 8√15)

Homework Equations


Numbers can be represented as √a - √b

The Attempt at a Solution


I had tried to make in an equation where the 2 expressions were equal, but after squaring both sides, I didn’t really know what to do.
I had also tried to use something like making a perfect square, and I did get a number, though it was much longer and more complicated, and I don’t think that’s right.
Now I’m a bit lost, and I’m not sure how to use the √a - √b that the teacher gave us.
Thanks!
 
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Cheesycheese213 said:

Homework Statement


Simplify √(53 - 8√15)

Homework Equations


Numbers can be represented as √a - √b

The Attempt at a Solution


I had tried to make in an equation where the 2 expressions were equal, but after squaring both sides, I didn’t really know what to do.
I had also tried to use something like making a perfect square, and I did get a number, though it was much longer and more complicated, and I don’t think that’s right.
Now I’m a bit lost, and I’m not sure how to use the √a - √b that the teacher gave us.
Thanks!
What did you get when you squared both sides? In particular, what did you get when you squared ##\ \sqrt{a}-\sqrt{b}\,? ##
 
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SammyS said:
What did you get when you squared both sides? In particular, what did you get when you squared ##\ \sqrt{a}-\sqrt{b}\,? ##
Ohhhh I see! Thank you so much!
 
Cheesycheese213 said:
Ohhhh I see! Thank you so much!
So, did you solve the problem? If so what is your answer?
 
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I got 4√3 - √5! Thanks again! :smile::smile::smile:
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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