<s>[INST] Finding the intersection of $x= y^2 + y - 2$ and $y = -x^2 - \frac32x + 1 $ Goodmorning, I am struggling in finding the points of intersection of the following parabolas:
$$\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$$
I know that these two can be solved either algebraically or with matrices, but at the moment the algebraic solution is the one that I'm going for.
How can I solve this system without getting lost in complex and boring calculations? (I'm asking specifically if there are some tricks to solve this more quickly end elegantly)
 [/INST] $\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$
The first one can be written as,
$x = (y-1) (y+2)$ ...(i)
The second one is $y = -x^2 - \frac32x + 1 \implies y-1 = - x^2 - \frac{3}{2}x$ ...(ii)
Plugging in $y-1$ from (ii) into (i),
$x = (- x^2 - \frac{3}{2}x) (- x^2 - \frac{3}{2}x + 3)$
$x = x (x + \frac{3}{2}) (x(x + \frac{3}{2})-3)$
$x = 0$ is a solution. If $x \ne 0$,
$x(x+\frac{3}{2})^2 - 3 (x + \frac{3}{2}) = 1$
$a^3 - \frac{3}{2} a^2 - 3 a - 1 = 0 \ $ where $a = x + \frac{3}{2}$
$a (a - 2)(a+\frac{1}{2}) - 2 (a+\frac{1}{2}) = 0$
$(a + \frac{1}{2})(a^2 - 2 a - 2) = 0$
So $a = -\frac{1}{2}, a = 1 \pm \sqrt3$
$\implies x = - 2, x = -\frac{1}{2} \pm \sqrt3$
So all in all, we got all $4$ values of $x$ at intersections,
$x = 0, - 2, -\frac{1}{2} \pm \sqrt3$. Find corresponding $y$ values.
</s>