Hello
I am trying to prove that -sqrt(5) + sqrt(11 + 6*sqrt(2)) = sqrt(2) + sqrt(14 - 6*sqrt(5)).
I don't even know where to begin.
Thanks a lot!
$\displaystyle -\sqrt{5} \ + \ \sqrt{11 + 6\sqrt{2}} \ \ \ \ =? \ \ \ \ \sqrt{2} \ + \ \sqrt{14 - 6\sqrt{5}}$
A possible way is to simplify $\displaystyle \ \sqrt{11 + 6\sqrt{2}} \ $ to the form $\displaystyle \ a + b\sqrt{2}, \ $ and to simplify $\displaystyle \ \sqrt{14 - 6\sqrt{5}} \ $ to the form $\displaystyle \ c + d\sqrt{5}, \ \ $
where a, b, c, and d are rational numbers.
$\displaystyle a + b\sqrt{2} \ = \ \sqrt{11 + 6\sqrt{2}} $
$\displaystyle (a + b\sqrt{2})^2 \ = \ (\sqrt{11 + 6\sqrt{2}} \ )^2 $
$\displaystyle a^2 + 2ab\sqrt{2} + 2b^2 \ = \ 11 + 6\sqrt{2} \ \implies$
$\displaystyle a^2 + 2b^2 \ = \ 11$
and
$\displaystyle 2ab\sqrt{2} \ = \ 6\sqrt{2}$
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Also,
$\displaystyle c + d\sqrt{5} \ = \ \sqrt{14 - 6\sqrt{5}} $
$\displaystyle (c + d\sqrt{5})^2 \ = \ (\sqrt{14 - 6\sqrt{5}} \ )^2 $
$\displaystyle c^2 + 2cd\sqrt{5} + 5d^2 \ = \ 14 - 6\sqrt{5} \ \implies$
$\displaystyle c^2 + 5d^2 \ = \ 14$
and
$\displaystyle 2cd\sqrt{5} \ = \ -6\sqrt{5}$
Hello!
Thank you very much for your reply. Unfortunately, I don't understand it at all. How does this turn the left side of the equation into the right one or vice versa?
Also, I don't understand how you conclude this:
a^2+2b^2 = 11
and
2ab = 6*sqrt(2)
And the respective step for the next equation. How did you arrive at this?
Thanks a lot!
My post (post #2) was being edited back and forth when you put it in a quote box. Please look at the most recent edits for the content to make the most sense.
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For example, the following you posted about was an error I was catching during my edit process:
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$\displaystyle a^2 + 2b^2 \ = \ 11$
and
$\displaystyle 2ab\sqrt{2} \ = \ 6\sqrt{2}$
Second equation:
Divide by $\displaystyle \ 2\sqrt{2} \ $ on each side:
ab = 3
Solve for either a or b in terms of the other, and substitute that into the first equation:
$\displaystyle a = \dfrac{3}{b}$
$\displaystyle \bigg(\dfrac{3}{b}\bigg)^2 \ + \ 2b^2 \ = \ 11$
$\displaystyle \dfrac{9}{b^2} \ + \ 2b^2 \ = \ 11$
$\displaystyle 9 + 2b^4 \ = \ 11b^2 $
$\displaystyle 2b^4 - 11b^2 + 9 \ = \ 0 $
Solve this for an appropriate value(s) for b, and then substitute back to find a. (It factors.)
Thanks a lot! Now I understand the part I mentioned. However, I still don't understand how this helps me.
Didn't you simply show that sqrt(11 + 6*sqrt(2)) is already in the form of a + b*sqrt(2)? Shouldn't I swap the 2 around, meaning show that sqrt(11 + 6*sqrt(2)) is in the form c + d*sqrt(5) and thus show that the left side is in the form of the right side? If so, wouldn't the algebra be significantly less clean? I don't think I could do it to be honest. I do suspect that the isolated square roots in either side of the equations would have something to do with it, right?
What happens if you do what I explain in my couple of posts is that you show that sqrt(11 + 6*sqrt(2)) = 3 + sqrt(2), and that sqrt(14 - 6*sqrt(5)) = 3 - sqrt(5).
Substitute these reduced quantities back into the sides of what you were given:
-sqrt(5) + (3 + sqrt(2)) $\displaystyle \ \ $vs. $\displaystyle \ \ $sqrt(2) + (3 - sqrt(5))
3 + sqrt(2) - sqrt(5) $\displaystyle \ = \ $ 3 + sqrt(2) - sqrt(5)
$11+6\sqrt{2} = 9 + 6\sqrt{2} + 2 = (3 + \sqrt{2})^2 \implies -\sqrt{5} + \sqrt{11+6\sqrt{2}} = \color{red}{-\sqrt{5} + (3 + \sqrt{2})}$
$14 - 6\sqrt{5} = 9 - 6\sqrt{5} + 5 = (3 - \sqrt{5})^2 \implies \sqrt{2} + \sqrt{14-6\sqrt{5}} = \color{red}{\sqrt{2} + (3 - \sqrt{5})}$
Hi guys! Sorry to ask for your help again but what about for this one? I've been struggling to apply the same reasoning:
sqrt(5 + 2*sqrt(5)) - sqrt(5 - 2*sqrt(5)) = sqrt(10 - 2*sqrt(5))
Thanks!
Edit: I tried squaring both sides and it worked, but what if it didn't?