Can someone help me to solve this equation for x and y?

$\displaystyle 5x^2+5y^2+8xy+2y-2x+2=0$

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- Apr 5th 2006, 02:31 PMOReillyEquation help
Can someone help me to solve this equation for x and y?

$\displaystyle 5x^2+5y^2+8xy+2y-2x+2=0$ - Apr 5th 2006, 08:59 PMearbothQuote:

Originally Posted by**OReilly**

the bad news first: I cann't show you how to solve this equation. I only can give you the solution, which I found by trial and error: x = 1; y = -1.

Your equation describes how to get the zeros of an elliptical paraboloid. The vertex of this curve is (1, -1, 0). I've attached a diagram to demonstrate my results.

Greetings

EB - Apr 6th 2006, 12:10 AMmathemagic
cool

- Apr 6th 2006, 02:03 AMOReillyQuote:

Originally Posted by**earboth**

- Apr 6th 2006, 03:05 AMRebesques
You cannot get an algebraic solution in the usual sense - every point (x,y) on the parabola is a solution.

Maybe you need to have something else, like y=y(x)? (wherever this is possible). - Apr 6th 2006, 04:24 AMtopsquark
(Sigh) Once again this morning, I am in a hurry.

As earboth said, there is no single point in general that will be a solution. However, unless I made a mistake (which is quite possible) I am getting that there are NO solutions.

Here's how to algebraically simplify the equation. You can check my Math to see if I was right.

At first glance the equation appears to be describing an ellipse that has been rotated about the origin. However there is a glitch...

First, we have an "xy" term in the equation we need to get rid of. Perform a coordinate rotation:

$\displaystyle x=x' cos\theta - y' sin \theta$

$\displaystyle y=x' sin \theta + y' cos \theta$

When you expand the terms (there are a lot of them!) look for the x'y' term. It will have a coefficient that depends on theta. Choose theta (I had pi/4) such that this coefficient is zero.

I had:

$\displaystyle 9x'^2+y'^2+2\sqrt2 y' + 2 = 0$.

We may complete the square on the y' variable and I got a form that reads:

$\displaystyle 9x'^2+(y'^2+\sqrt2 /3)^2 + 28/36 = 0$

This LOOKS like an ellipse, but note that the constant term here is positive. This implies that the axes of the ellispse are imaginary. Hence there is no (real) solution for x' and y' and thus none for x and y.

Again, please check the Math because I did this quickly. At the very least this post gives a way to simplify the original equation.

-Dan - Apr 6th 2006, 04:51 AMearbothQuote:

Originally Posted by**topsquark**

that's really brillant!

At your last step: You didn't realized, that $\displaystyle y'^2+2\sqrt2 y' + 2 $ is a complete square already. Now your equation reads:

$\displaystyle (3x')^2+(y'+\sqrt2)^2 = 0$.

So you have a sum of two squares which should be zero. It isn't very probable.

Greetings

EB - Apr 6th 2006, 07:09 AMThePerfectHacker
Do this,

$\displaystyle 5x^2+5y^2+8xy+2y-2x+2=0$

Express as,

$\displaystyle 5y^2+y(8x+2)+(5x^2-2x+2)=0$

Thus,

$\displaystyle y=\frac{-8x\pm \sqrt{-36x^2+72x-36}}{10}$

Thus,

$\displaystyle y=\frac{-8x\pm \sqrt{-36(x-1)^2}}{10}$.

Which has**real**solutions where,

$\displaystyle -36(x-1)^2\geq 0$, thus,

$\displaystyle (x-1)^2\leq 0$ which only happens when,

$\displaystyle x=1$

From here we can solve for $\displaystyle y$,

$\displaystyle y=-1$ - Apr 6th 2006, 10:21 AMtopsquark
Ah well. I DID suspect I had screwed up somewhere! :)

-Dan - Apr 6th 2006, 11:20 AMDenMac21
This is the best solution in my opinion and most elegant.

$\displaystyle \begin{array}{l}

5x^2 + 5y^2 + 8xy + 2y - 2x + 2 = 0 \\

4x^2 + 4y^2 + 8xy + x^2 + y^2 + 2y - 2x + 1 + 1 = 0 \\

(2x + 2y)^2 + (x - 1)^2 + (y + 1)^2 = 0 \\

\end{array}

$

Now, we have to solve system of equations:

$\displaystyle \begin{array}{l}

x + y = 0 \\

x - 1 = 0 \\

y + 1 = 0 \\

\end{array}

$

which gives us that solutions are: $\displaystyle x = 1 \wedge y = - 1$ - Apr 6th 2006, 11:32 AMThePerfectHackerQuote:

Originally Posted by**DenMac21**

- Apr 6th 2006, 01:28 PMtopsquark
Well, I think MY solution was the best. Even though it, umm... didn't work. :eek:

-Dan

(It could've, though!)