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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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As usual with me I can't find an efficient (or in this case even elegant) way to do this. But I had an idea...

\(\displaystyle \dfrac{2xy}{x + y} + \sqrt{ \dfrac{x^2 + y^2}{2} } = \sqrt{xy} + \dfrac{x + y}{2}\)

Let a = y/x.

Two points:

1) Watch out for x = 0 later.

2) I tacitly multiplied the first fraction by a/a. So watch out of a = 0, as well.

Then, after a bit

\(\displaystyle \dfrac{2a}{a + 1} + \dfrac{1}{ \sqrt{2} } ~ \sqrt{a^2 + 1} = \sqrt{a} + \dfrac{1}{2} (a + 1)\)

Isolate the \(\displaystyle \sqrt{a + 1}\) and square. After some more simplifying and clearing the fractions

\(\displaystyle 2 (a^2 + 1)(a + 1)^2 = 16 a^2 + 4 a(a + 1)^2 + (a + 1)^4 - 8 a (a + 1)^2 + ( -16 a(a + 1) + 4 a (a + 1)^3 ) \sqrt{a}\)

Expanding and simplifying

\(\displaystyle 2a^4 + 4a^3 + 4a^2 + 4a + 2 = (a^4 + 14a^2 + 1) + (4a^3 - 4a^2 - 4a + 4) \sqrt{a}\)

\(\displaystyle a^4+ 4a^3 - 10a^2 + 4a + 1 = (4a^3 - 4a^2 - 4a + 4) \sqrt{a}\)

So far my approach for finding a is pretty standard. Isolate one square root, simplify, then the next step would be to isolate the other square root and simplify. But there is an issue with this procedure: If a = 1 (as it will be when we solve it) then we have both sides of this equal to 0! I have chosen to change the variable to \(\displaystyle b^2 = a\). This means we don't have to square out that LHS, which is ugly enough. But again, we have to be careful about values of b now. (Don't worry, it'll all come together.)

So with the substitution and (yet more) simplifying:

\(\displaystyle b^8 - 4 b^7 + 4 b^6 + 4 b^5 - 10 b^4 + 4 b^3 + 4 b^2 - 4 b + 1 = 0\)

The rational root theorem says that \(\displaystyle b = \pm 1\) are the only rational solutions. Instead of some fancy theorems to prove these are the only real solutions, I simply graphed it. It was faster.

So, \(\displaystyle b = \pm 1\) are the only real solutions. Time to work backwards. \(\displaystyle a = b^2 = 1\). Thus \(\displaystyle \dfrac{y}{x} = 1\) and we get that the solutions to the original equation are all y = x. What about the x = 0 thing? y = x = 0 violates the original equation.

So the final solution is all real \(\displaystyle y = x, ~ x \neq 0\).

-Dan

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Well done, topsquark ! You know, I sense your obsession lately with my POTWs and challenge problems, hehehe...I hope so far you found nothing but fun in tackling all those problems!

It's not so much that I've been obsessed, it's more that I have actually been able to solve some recently. I usually try to work them out.

-Dan

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It's not so much that I've been obsessed, it's more that I have actually been able to solve some recently. I usually try to work them out.

-Dan

(Whispers) Why are we having a conversation using spoilers?

-Dan

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Because it is fun!