# Solving Quadratic Eqn for Symbolic Complex Eqn

1. Mar 8, 2013

### WesleyJA81

1. The problem statement, all variables and given/known data

Hey guys, I know there are people on here that can help me figure this out...

I have a complex quadratic. The coefficients of the quadratic are a=(1-x) b=x*(u+P/(u*r)) and c=-(u^2 + x*P/r). Using -b/2a +/- sqrt(b^2-4*a*c)/2a

I'm trying to analytically reduce this quad eqn as far as I can

I know this should reduce to 1/(x-1)*u + x*P/(r*u)

3. The attempt at a solution

As far as I could take it was trying to reduce the term in the sqrt => b^2-4*a*c

u^2*(x-2)^2 + (P/(u*r))^2*(x)^2 + P/r*(-4x^2 + 4x)

I feel like I should be able to factor this into something like

(u+p/(u*r))^2 * (something)^2 so I can take the sqrt but I can't figure this out. Thanks in advance for the help.

~ Wesley

Last edited: Mar 8, 2013
2. Mar 8, 2013

### eumyang

Don't write this. Leave it in the "unfactored" form for a moment. There should be 6 terms, three of which are perfect squares. It will be in the form of
$a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = (a + b + c)^2$

See if you can get the polynomial in the above form.

(Mods: I hope that this isn't too much of help. If so, please delete.)

3. Mar 8, 2013

### WesleyJA81

eumyang, thanks for the tip. Let me take a look. The problem I'm working on is bigger than what I've posted. This is just the point that has stumped me. Thanks again!

Last edited: Mar 8, 2013
4. Mar 8, 2013

### WesleyJA81

It ended up being of the form

$a^2 + b^2 + c^2 - 2ab + 2bc - 2ac = (-a + b + c)^2$

I'm embarrassed at the amount of time I spent trying to figure that out. Thanks again