Solving a Quadratic Equation with Two Variables

[Ashley1nOnly]

Homework Statement

X^2-yx+y^2-7=0

Homework Equations

-b +- sqrt(b^2-4ac)/2a

The Attempt at a Solution

Trying to complete the square with two variables
(x^2-yx)+(y^2)=7

(X^2-yx+y/2)+y^2=7
Where else from here. I'm just having problems because of the y variable

 

[Mark44]

Homework Statement

X^2-yx+y^2-7=0

What's the complete problem statement? IOW, what are you supposed to do here?

Homework Equations

-b +- sqrt(b^2-4ac)/2a

The Attempt at a Solution

Trying to complete the square with two variables
(x^2-yx)+(y^2)=7

(X^2-yx+y/2)+y^2=7
Where else from here. I'm just having problems because of the y variable

 

[Ashley1nOnly]

The original equation was a differential equation
(2x-y)dx+(2y-X)dy= 0
Which I solved and got
X^2-yx+y^2=7

It took the form of
M(X,y)dx+N(X,y)=0

M(partial y)= N(partial X)

So it had an exact solution
I then continued the rest of the steps and got my final answer but it needs to be in terms of y.

The final answer is

Y=[x+sqrt(28-3x^2)]/2, |x|<sqrt(28/3)

I'm trying to transform my answer into their final answer.

 

[Ashley1nOnly]

I need help completing the squares so that I can use the quadratic equation to solve for y.

 

[ehild]

I need help completing the squares so that I can use the quadratic equation to solve for y.

You need y as function of x. So write y²-yx+x²-7 in form (y-x/2)²+f(x)

 

[Mark44]

@Ashley1nOnly, your first post should have started like this:

Homework Statement

Solve for y in the equation X^2-yx+y^2-7=0

Homework Equations

-b +- sqrt(b^2-4ac)/2a

One other thing -- what you wrote above would be interpreted by most as
$$-b \pm \frac{\sqrt{b^2 - 4ac}}{2} a$$

 

[Ray Vickson]

Homework Statement

X^2-yx+y^2-7=0

Homework Equations

-b +- sqrt(b^2-4ac)/2a

The Attempt at a Solution

Trying to complete the square with two variables
(x^2-yx)+(y^2)=7

(X^2-yx+y/2)+y^2=7
Where else from here. I'm just having problems because of the y variable

You have an equation of the form $y^2 + b y + c = 0$, with appropriate $b$ and $c$. The solution forms
$$y_1 = \frac{-b + \sqrt{b^2 - 4 c}}{2} , \; y_2 = \frac{-b - \sqrt{b^2 - 4 c}}{2} $$
will give you the answer. It does not matter if $b,c$ are numerical constants, or if they are 1000-page formulas containing 500 other variables; as long as they do not contain $y$, the quadratic solution formula holds and you are good to go.