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by **Mumina** on Sat Sep 05, 2009 12:47 pm

This is the beginning of our algebra course and I am already lost! This seems so simple, solve for x in terms of y:

4x^2-2xy+3y^2=2

I can't figure out how to isolate x on one side. The y keeps popping up! Help please!

**Sponsor**

by **stapel_eliz** on Sat Sep 05, 2009 1:43 pm

Mumina wrote:...solve for x in terms of y: 4x^2-2xy+3y^2=2

The trick here is to view this as a quadratic in y:

. . . . .3y² - 2xy + 4x² = 2

You can apply the Quadratic Formula to the equation, with a = 3, b = -2x, and c = 4x².

. . . . . $y\, =\, \frac{-(-2x)\, \pm\, \sqrt{(-2x)^2\, -\, 4(3)(4x^2)}}{2(3)}$

...and so forth.

You will get "y=" two different things, one from the "plus" and the other from the "minus" of the "plus / minus" in the Formula. :wink:

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solve 4x^2-2xy+3y^2=2 for x in terms of y

solve 4x^2-2xy+3y^2=2 for x in terms of y