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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?

User
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DSL
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2 Answers2

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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.

Ennar
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  • Not necessarily, It can have at most 9 Solutions. – Ashar Tafhim Sep 01 '16 at 13:28
  • @AsharTafhim, yes, you are right. Thank you. – Ennar Sep 01 '16 at 13:29
  • I solve x^4-4x^2+1, and got {-0.517638, 0.517638, -1.93185, 1.93185}. and I make a copy of this list of solution and reverse the order to get {1.93185, -1.93185, 0.517638, -0.517638}. then transpose the two lists I got the x and y coordinates of the other points.{{-0.517638, 1.93185}, {0.517638, -1.93185}, {-1.93185, 0.517638}, {1.93185, -0.517638}}. However, I still a little disoriented and do not full understand why this works. Would you please illustrate a little more on this? – DSL Sep 01 '16 at 16:00
  • @ Ennar By the way, by expanding the system, I got 16 x - 68 x^3 + 48 x^5 - 12 x^7 + x^9 instead of 15 x - 68 x^3 + 48 x^5 - 12 x^7 + x^9. Why is that? – DSL Sep 01 '16 at 16:10
  • @Tmm it's because you forgot to subtract the $x$ from the other side. I will write down more details. – Ennar Sep 01 '16 at 16:19
  • @Tmm, please see my edit and let me know if this is clear enough now. – Ennar Sep 01 '16 at 16:45
  • @Ennar, Thank you so much for your effort, it certainly makes is very clear to me now. By the way, I use WolframAlpha to expand (x^3 - 4 x)^3 - 4 (x^3 - 4 x) and it still givesx^9-12 x^7+48 x^5-68 x^3+16 x Please check this and let me know what went wrong. http://www.wolframalpha.com/input/?i=Expand%5B(x%5E3+-+4+x)%5E3+-+4+(x%5E3+-+4+x)%5D – DSL Sep 01 '16 at 19:18
  • @ Ashar Tafhim, would you mind giving more details of your solution? thank you. specifically "the where x is cube root of +2

    or −2

    . "

     – DSL Sep 01 '16 at 19:22
  • @Tmm, you are welcome. Expanding is correct, but you want to solve $(x^3 - 4 x)^3 - 4 (x^3 - 4 x) = x$, not $(x^3 - 4 x)^3 - 4 (x^3 - 4 x) = 0$. – Ennar Sep 01 '16 at 19:22
  • @Ennar thanks. What subjects in math are specifically dealing with solving equations like this Algebraically? Besides Calculus. Perhaps you could recommend a book or name of the subject? I am just a beginner. Thanks. – DSL Sep 01 '16 at 19:27
  • @Tmm I have eedited my answer. – Ashar Tafhim Sep 01 '16 at 20:33
  • @Tmm, Algebra deals with algebraic equations like these, but algebraic equations are also studied by Algebraic Geometry. I'm not sure what books in english are appropriate at beginner level, but Algebra is done in great detail in books by Hungerford and Lang (both named Algebra). I wouldn't recommend Lang at beginner level, and probably the best way to go would be to find some lecture notes on introductory algebra courses available online. – Ennar Sep 02 '16 at 09:56
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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

Ashar Tafhim
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