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Solve the quadratic equation in y for y in terms of x

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by budoyong » Sun Apr 19, 2009 11:20 pm
Solve the quadratic equation in y for y in terms of x, and express y as two functions of x.

x³-x²+2xy-y²=0, Solution is: {y=(1+√x)x},{y=(1-√x)x}

can somebody help me how did this happen, thanks
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by DanaJ » Mon Apr 20, 2009 12:00 am
It's actually quite simple if you know where to start from.

First off, just re-arrange this equation to emphasize the fact that it's a quadratic equation in y:

-y^2 + y*2x + x^3 - x^2 = 0

I usually don't like to start with that minus in front of y^2, so you get:

y^2 - y*2x + x^2 - x^3 = 0.

The discriminant will be 4x^2 - 4(x^2 - x^3) = 4x^3.

The solutions will be
1. [2x - sqrt(4x^3)]/2 = [2x - 2xsqrt(x)]/2 = x - x*sqrt(x) = x(1 - sqrt(x))

2. [2x + sqrt(4x^3)]/2 = [2x + 2xsqrt(x)]/2 = x + x*sqrt(x)= x(1 + sqrt(x))
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by truplayer256 » Mon Apr 20, 2009 11:59 am
Dana explained the solution of this problem very well. Here's how I solved the problem:

x^3-x^2+2xy-y^2=0--> This is the same thing as x^3-(x^2-2xy+y^2)=0
and we know that x^2-2xy+y^2 is equal to (x-y)^(2). So, from this, it's safe to assume that x^3-(x-y)^(2)=0 and now you can just solve this as if it were an algebraic expression:

x^3-(x-y)^(2)=0
x^3=(x-y)^(2)
Take the square root of both sides:

+x^(3/2)=x-y
-x^(3/2)=x-y

x-x^(3/2)=y and x^(3/2)+x=y
x(1-x^(1/2)=y x(x^(1/2)+1)=y
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by dumb.doofus » Mon Apr 20, 2009 12:08 pm
Treat that as a quadratic equation for y.. ay^2 + by + c = 0 we already know the roots for a quadratic equation.. just put the values..

a = 1,
b = -2x
c = x^2 - x^3
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