Solving 2y² + xy = x² + 3 for y

  • Thread starter Thread starter Jeff Ford
  • Start date Start date
  • Tags Tags
    Stuck
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 3K views
Jeff Ford
Messages
154
Reaction score
2
I'm trying to solve the following equation for y
[tex]2y^2 + xy = x^2 + 3[/tex]
So far I've gotten it down to [tex]y = \frac{x^2 + 3}{2y = x}[/tex]
Or I've tried [tex](2y-x)(y+x) = 3[/tex]
But I'm stuck at that. Any advice would be appreciated.
 
Physics news on Phys.org
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!
 
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.
 
So it would be
[tex]y = \frac{-x \pm \sqrt{9x^2 + 24} }{4}[/tex]
 
Exactly, as you see: you can use it with variables as well :smile: