Solve using quadratic formula?

[Kinetica]

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

Hey guys, recently, I struggled to solve this equation.
I need to find θ1 and θ2 by using this system of equations:

θ1+θ2=γ
θ1*θ2=β

After letting θ2=β/θ1, I plugged this into the first equation. I got:

θ12-γθ1+β=0

At this point I get a very unsexy θ1 result by using quadratic solution.

Need you help!

P.S. This way, I get 4 solutions overall, but my professor is saying that there should only be 2 solutions.

[Dick]

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

Hey guys, recently, I struggled to solve this equation.
I need to find θ1 and θ2 by using this system of equations:

θ1+θ2=γ
θ1*θ2=β

After letting θ2=β/θ1, I plugged this into the first equation. I got:

θ12-γθ1+β=0

At this point I get a very unsexy θ1 result by using quadratic solution.

Need you help!

What you've done so far is just fine. And I don't think the answer is going to pretty no matter how you do it.

[Ray Vickson]

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

Hey guys, recently, I struggled to solve this equation.
I need to find θ1 and θ2 by using this system of equations:

θ1+θ2=γ
θ1*θ2=β

After letting θ2=β/θ1, I plugged this into the first equation. I got:

θ12-γθ1+β=0

At this point I get a very unsexy θ1 result by using quadratic solution.

Need you help!

P.S. This way, I get 4 solutions overall, but my professor is saying that there should only be 2 solutions.

How did you get 4 solutions? Could you show us? The point is that you must have made an error, and we cannot help until we know what you did.

RGV

[epenguin]

If you look at your equations they are symmetric between θ1 and θ2. You could say there is no difference between θ1 and θ2. Or, do what you did to get a quadratic equation for θ1 but do it to get an equation for θ2 instead - you find you have got exactly the same equation for θ2 as you got for θ1.

You haven't really got two quadratic equations, you've got one.