Real Numbers and Equality: Can x+y=4 Be Proven from x2-4x=y2-4y?

crakajackxx

Homework Statement



x and y are real numbers. prove that if x2-4x=y2-4y and x not equal to y, then x+y=4.

Homework Equations



n/a

The Attempt at a Solution



I tried using cases and making x and y positive and negative or even and odd, and that didnt work. then i tried completing the square on both sides and I got x=y.
please help~
 
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Move all the terms to one side. x^2-y^2=(x+y)(x-y). Factor out an (x-y).
 
Dick said:
Move all the terms to one side. x^2-y^2=(x+y)(x-y). Factor out an (x-y).
Dick, I think you mean "move all the squared terms to one side."
 
Mark44 said:
Dick, I think you mean "move all the squared terms to one side."

I meant what I said. I was counting on the OP finish it from there.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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