1. Jun 19, 2012

Kartik.

Taking a general function with two quadratic eqs,
Y = (ax^2+bx+c / px^2+qx+r)
ax+bx+c = pyx^2+qxy+ry
x^2(a-py) + x(b-qy)+(c-ry)=0
The discriminant turns out to be a+ve or a 0.
So,
(b-qy)^2 - 4(a-py)(c-ry) >or= 0
Now how does this equation yield a range for y?or can it?what are the next steps?

2. Jun 19, 2012

HallsofIvy

Staff Emeritus
You say you have "two quadratic equations" but I see three equations, none of which is quadratic.

3. Jun 20, 2012

Kartik.

that is probably because of me using the m.physicsforum because of which i lack a "^2”(at the 2nd equation, the way you put it) and a couple of signs of implications after the first equation(which is the attempt towards the solution) and by saying i have two quadratic equations, i mean to say one quadratic eq over the other = quadratic fraction