How do you solve in terms of y 4x^2-2xy+3y^2=2

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4x^2-2xy=2-3y^2
x(4x-2y)=2-3y^2 then I'm stuck[/B]
 

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  • #2
Delta2
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View the equation as a quadratic ##ay^2+by+c=0## with ##a=3, b=-2x, c=4x^2##, that is if you want to find ##y=f(x)##.
if you want to find ##x=f(y)## then view it as a quadratic ##ax^2+bx+c=0## with ##a=4,b=-2y, c=3y^2##.
 
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What do I do with the 2 at the end of the original equation?
 
  • #4
SammyS
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Homework Statement




Homework Equations




The Attempt at a Solution


4x^2-2xy=2-3y^2
x(4x-2y)=2-3y^2 then I'm stuck[/B]
Please include a complete statement of your problem in the text of the post itself, no matter what you state in the title.
 
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  • #5
Delta2
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What do I do with the 2 at the end of the original equation?
Ah yes sorry, I thought it was =0, just incorporate 2 inside c, so ##c=4x^2-2## (or ##c=3y^2-2##)
 
  • #6
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Ah! Well that's going to be fun! Thank you so much!
 

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