Solve in $\mathbb{R}$: $(x^2-3x-2)^2-3(x^2-3x-2)-2-x=0$

I'm supposed to solve this equation. It's from a math contest so solving it by hand would be preferable (no quartic formulas). I thought about making $u = x^2-3x-2$ obviously but it leads to another quartic equation. I also tried the substitution $u=x+2$, and after the whole expand trinomial, simplify, invoke rational root theorem and test roots, I still got nothing out of it.

I noticed that $x^2-3x-2$ can't be factored nicely so I dunno what other route to take. Lots of equations I tackled in math contests could make use of nice trigonometric substitutions, but none in particular pop in my head right now.

If anyone can give me hints or a full solution, that would be awesome. Thanks!

  • 1
    $\begingroup$ Is the math contest this problem was taken from already finished? Or is it still on-going? $\endgroup$ – Joel Reyes Noche Aug 6 '14 at 1:34
  • 5
    $\begingroup$ So $f(f(x))=x$, interesting. $\endgroup$ – blue Aug 6 '14 at 1:38
  • 1
    $\begingroup$ Try solving $f(x) = x$ with $f(x) = x^2 -3x -2$. What is $f(f(x))$ then equals? $\endgroup$ – Winther Aug 6 '14 at 1:40
  • 1
    $\begingroup$ @JoelReyesNoche This is a strange question. I don't think that if I was cheating I'd admit it. Anyway it's from a pretty old contest: obm.org.br/export/sites/default/revista_eureka/docs/… page 49, problem 74. $\endgroup$ – Deathkamp Drone Aug 6 '14 at 1:45
  • 2
    $\begingroup$ @DeathkampDrone Please don't take offense at Joel's question--I've seen a lot of live contest problems asked on this site, and quite a few people freely admit they're asking for help on them. $\endgroup$ – apnorton Aug 6 '14 at 1:46

Since you have a quartic, there are potentially 4 real solutions.

Note that any solution to $x = x^2 - 3x - 2$ is also a solution to the given equation.

Hence, $x^2 - 4x - 2$ is a factor of the quartic. Now find the other factor, and solve both quadratics.

Alternatively, see this almost 10 year old thread on the Art of Problem Solving Forum, or this slightly newer thread which discusses a similar problem.

| cite | improve this answer | |
  • $\begingroup$ +1 This is an extremely good answer. I wish more people on this site would answer like this. $\endgroup$ – Joel Reyes Noche Aug 6 '14 at 1:49
  • $\begingroup$ I don't understand why the second line is true, could you give some more detail? blue's comment notes that $f(x)=x^2-3x-2\Rightarrow f(f(x))=x$, something that I didn't notice the first time. Are you using this fact to deduce the second line? I think I'm missing something completely obvious... $\endgroup$ – Deathkamp Drone Aug 6 '14 at 1:59
  • $\begingroup$ If $x = x^2-3x-2$, then substitute $x^2-3x-2$ for $x$ on the right side to get $x = (x^2-3x-2)^2-3(x^2-3x-2)-2$, which is the given equation. Hence, the solutions to $x = x^2-3x-2$ are also solutions to the given equation. $\endgroup$ – JimmyK4542 Aug 6 '14 at 2:04
  • 1
    $\begingroup$ Oh! I just stared at this, confused, for 5 minutes or so. Normally I substitute $y=f(x)$ but I had to think a little more to get the $x=f(x)$ substitution. I understand it now, and I can finish the problem. Thank you for your help! $\endgroup$ – Deathkamp Drone Aug 6 '14 at 2:14
  • $\begingroup$ (Also, thanks for taking the time to find these threads. Lots of insightful answers.) $\endgroup$ – Deathkamp Drone Aug 6 '14 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.