I have the following system: $$\begin{cases} ax^2-2x+y^2-2=0 \\ -3x^2+x-2y+2=0 \end{cases}$$
I need to find all $a\in \mathbb{R}$ for which the system above has 2 real solutions. The problem also asks me to solve the system for $a=1$. It looks like the system has 2 real solutions and 2 complex solutions for $a=1$.
Substituting the $y$ from the second equation into the first one leads to a 4th degree polynomial which I can't seem to be able to solve.
I also tried completing squares in a bunch of ways with no succes.
I have noticed that the $x$ terms from the second equation can be written as $-3x^2+x-2=-(3x+2)(x-1)$. I doubt this can be used in any meaningful way.
Any ideas on how to solve this? Thanks for your help!
Write $y=\frac{1}{2}(-3x^2+x+2)$ by the second equation. Then the first equation is $$ 4ax^2 + 9x^4 - 6x^3 - 11x^2 - 4x - 4=0 $$ For $a=1$ this equation has two complex solutions and two real solutions - as you said. In general, for a polynomial, we can use the rule of Descartes and others.
References:
Number of real roots of a polynomial