$Q_{30}$ Graph the curves $y = x^3-4x$ and $x=y^3-4y$ and find their points of intersection correct to one decimal place.

Question

I could not find all the points of intersection

Here is what I did.

First, I use elimination to obtain $(x_1,y_1)$, and $(x_2,y_2)$.

For $y=x^3-4x$, I set $x_1 = t$, then $y_1=t^3-4t$.

For $x=y^3-4y$, I set $y_2 = t$, then $x_2=t^3-4t$.

Then I plot the equation parametrically as follow

check plotting of the equations

Then I equal $x_1$ to $x_2$, $y_1$ to $y_2$ and I got

$$t^3 - 4 t - t =0$$

solve for $t$ I got $3$ points $\{(0, 0), (-\sqrt{5}, -\sqrt{5}), (\sqrt{5}, \sqrt{5})\}$.

check plot with all points of intersection

But there are more than 3 points of intersection, how do I get the other points?

Answer 1

You just have to solve system $$ y=x^3-4x$$ $$x = y^3-4y $$ by substitution. You get a 9th degree polynomial, and thus there are at most 9 intersections.

By substituting we get $$x = (x^3-4x)^3-4(x^3-4x) \implies \underbrace{x^9-12 x^7+48 x^5-68 x^3+15 x}_{=: p(x)} = 0$$

Thus, if $(\alpha,\beta)$ is a solution of the system, then $\alpha$ is root of $p$. Conversely, if $\alpha$ is root of $p$, then $(\alpha,\alpha^3-4\alpha)$ is a solution of the system.

We can't find roots of 9th degree polynomial in general, so we need to "guess" some roots. What we will do is find some solutions of the above system and they will give us some of the roots of $p$.

Notice that you already found some solutions: $\{(0,0),\pm (\sqrt 5,\sqrt 5)\}$ by assuming that they are of form $(t,t)$. But, if $(t,t)$ is a solution, $t$ must be a root of $x(x^2-5)$. Since $t$ is root of $p$ as well, it means that $x(x^2-5)$ divides $p(x)$.

Similarly, it is easily verified that there are solutions of the form $(t,-t)$ as well, and they are roots of $x(x^2-3)$, and thus $x(x^2-3)$ divides $p(x)$.

Hence, we conclude that $$x^9-12 x^7+48 x^5-68 x^3+15 x = x(x^2-3)(x^2-5)q(x)$$ Use long division to find that $q(x) = x^4 - 4x^2 +1$. Now you can find all solutions.

Answer 2

You found the solution when $x=y$.For $x=-y$ ,we get two new solutions,where $x=\sqrt 3$ or $x=-\sqrt 3$ simply by putting $x=-y$. So we now have 4 solutions.Now suppose $x\neq y$ and $x\neq-y$. Then add the given two equations and divide by $x+y$ to get $5=x^2 -xy +y^2 $.Similarly, subtract these two equations and divide by $x-y$ to get $ 3=x^2 +xy +y^2$.Manipulate these two equations to get $xy=-1$ and $ x^2 +y^2 =4$.Here are the results.

https://www.wolframalpha.com/input/?i=intersection+of+xy%3D-1+and+x%5E2+%2By%5E2+%3D4