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)
 A: If, in the equality $x=y^2+y-2$, you replace $y$ with $-x^2-\frac32x+1$, then you will get$$-x^4-3 x^3+\frac34x^2+\frac{11}2x.\tag1$$It is clear that $x=0$ is a solution (and then $y=1$). If you divide $(1)$ by $x$ and multiply it by $-4$, you will get$$4 x^3+12 x^2-3 x-22\tag2$$and then the rational root theorem allows you to see that $-2$ is a solution (and then $y=0$). And if you divide $(2)$ by $x+2$, you will get $4 x^2+4 x-11$. So, all that remains to be done is to find the roots of this quadratic equation.
A: $$x = y^2 + y - 2 \tag 1$$
$$y = -x^2 - \frac32x + 1 \tag 2$$
Inspection :
$x=0$ in Eq.(2) gives $y=1$ and $y=1$ in Eq.(1) gives $x=0$ . Thus a first solution is
$$(x=0\:;\:y=1)$$
$y=0$ in Eq.(1) gives $x=-2$ and $x=-2$ in Eq.(2) gives $y=0$ . Thus a second solution is
$$(x=-2\:;\:y=0)$$
Full solving :
Put $y$ from Eq.(2) into Eq.(1) :
$$x = (-x^2 - \frac32x + 1)^2 + (-x^2 - \frac32x + 1) - 2 $$
$$x^4+3x^3-\frac34 x^2-\frac{11}{2}x=0$$
Knowing the two above roots it is easy to factor :
$$x(x+2)(4x^2+4x-11)=0$$
Solving $4x^2+4x-11)=0$ leads to the two other roots. for each one Eq.(1) gives $y$.
The third solution is :
$$x=-\frac12+\sqrt{3}\quad ; \quad y=-\frac32-\frac12\sqrt{3}$$
The fourth solution is :
$$x=-\frac12-\sqrt{3}\quad ; \quad y=-\frac32+\frac12\sqrt{3}$$
A: $\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$
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.
