Parabola and Hyperbola Intersection I have been studying the intersection of geometric shapes lately, and I have stumbled upon this question that I just couldn't do. It describes two equations, one for a parabola and one for a hyperbola ($y=x^2 -3$ and $y = 2/x$).
I have learned techniques for solving these such as using simultaneous methods, however, when I used simultaneous, I came across this nasty thing that I couldn't solve ( a limitation of my algebra??): $0 = x^3 - 3x - 2$
For the life of me, I can't seem to figure out a way to a) simplify that equation or b) find a better way to solve the intersection itself
So far, I can see that there very much is a very elegant solution as when graphed, one can see that they intersect very nicely.
Thank you very much!!
 A: By inspection,
$$x^3-3x-2$$ has the root $x=-1$. Then by division,
$$x^3-3x-2=(x+1)(x^2-x-2).$$
The rest is easy.

Alternatively:
Your cubic equation is of the "deflated" type (no quadratic term). You solve such equations by setting $x:=u+v$, so that
$$x^3-3x-2=u^3+3uv(u+v)+v^3-3(u+v)-2.$$
Now if you choose $u$ and $v$ such that $uv=1$, the equation simplifies to
$$u^3+v^3-2=0.$$
Multiplying by $u^3$, you get
$$u^6-2u^3+u^3v^3=u^6-2u^3+1=(u^3-1)^2=0.$$
This gives the solution $u=1$, hence $x=2$, which is right.
To get the other roots, you have two options


*

*use complex numbers and compute the complex cubic roots of $1$,

*divide the initial polynomial by $x-2$ and solve the resulting quadratic, $x^2+2x+1$.
A: There is no straightforward simple way to solve a third-degree equation. There are methods and even a formula (if you're curious google for Cardano method or Cardano cubic formula), but they are not practical at all.
However, you can easily test some roots. From Rational Root Theorem we know that any rational root of the equation is if the form $x'=p/q $, where $p $ is a divisor of $-2$ (the independent term) and $q $ is a divisor of $1$ (the highest-degree coeficient).
Then, you know $$ x' \in \{ \pm 1, \pm 2 \}$$.
If you test these numbers you'll find at least one root.
Then by polynomial division you get a quadratic, which I think you can solve.
