Solving 2y^2 + xy = x^2 + 3: Stuck at y = \frac{x^2 + 3}{2y = x}

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Homework Help Overview

The discussion revolves around solving the equation 2y² + xy = x² + 3 for the variable y. Participants are exploring the nature of the equation, which is quadratic in y, and are considering different approaches to isolate y.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation into a form that isolates y, considering both a direct approach and a factorization method. Some participants suggest treating y as a variable and x as a constant, indicating that completing the square could be a viable strategy. Others note the implications of the quadratic nature of the equation, highlighting that a unique solution for y may not exist.

Discussion Status

Participants have provided various insights into the problem, with some suggesting methods for solving the quadratic equation and others confirming the presence of multiple solutions. There is an acknowledgment of the need for a "plus-minus" symbol in the final answer, indicating a productive direction in the discussion.

Contextual Notes

Participants are discussing the implications of having variables in the coefficients of the quadratic equation, which adds complexity to the solution process. The original poster's attempts reflect the constraints of the problem as a homework exercise.

Jeff Ford
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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.
 
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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]
 
Yes, that's exactly it.
 
Much obliged!
 
Exactly, as you see: you can use it with variables as well :smile:
 

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