Solving for y

[Jeff Ford]

I'm trying to solve the following equation for y
2y^2 + xy = x^2 + 3
So far I've gotten it down to y = \frac{x^2 + 3}{2y = x}
Or I've tried (2y-x)(y+x) = 3
But I'm stuck at that. Any advice would be appreciated.

[pizzasky]

Well for this sort of thing, I think we need to treat y as a variable and x as a constant. This leaves us with a quadratic expression in y (with x^2 + 3 as a constant), which means that we can complete the square and make y the subject of the equation...

Final tip: In your final answer, there should be a "plus-minus" symbol somewhere...

All the best!

[TD]

You won't be able to get a unique solution for y, since the equation is quadratic in y. If you see that, you can just solve it like any other quadratic equation, using the abc-formula. The only difference is that there won't only be numerical coefficients, but also x's but that's no problem.

[Jeff Ford]

So it would be
y = \frac{-x \pm \sqrt{9x^2 + 24} }{4}

[HallsofIvy]

Yes, that's exactly it.

[Jeff Ford]

Much obliged!

[TD]

Exactly, as you see: you can use it with variables as well :smile: