21
$\begingroup$

I have been trying to solve this one problem from the Duke Math Meet, which does not provide a solution:

Find all solutions of $(x^2+7x+6)^2 + 7(x^2+7x+6) + 6=x$

At first I tried to factorize the polynomial, but always had the right hand side x remain, which was inconvenient. I then tried using the fact that one can write the left hand side as a composition of functions, and equated that with the inverse of the quadratic plugged into the function, but that was a very nasty equation with square roots, still ending up with a quartic.

What other solution paths are viable for this problem, and is there a way to factor the quartic?

$\endgroup$
  • $\begingroup$ Do you mean $=0$ ? $\endgroup$ – Rene Schipperus Mar 7 '18 at 1:15
  • $\begingroup$ @ReneSchipperus Pretty sure no, he/she mentions "keeping $x$ on the right hand side" in his steps. $\endgroup$ – user535339 Mar 7 '18 at 1:16
  • $\begingroup$ @idk it says "finding the roots" in the title? $\endgroup$ – Andrew Li Mar 7 '18 at 1:19
  • $\begingroup$ IDK about it - idk $\endgroup$ – user535339 Mar 7 '18 at 1:21
  • 11
    $\begingroup$ This might be interesting: math.stackexchange.com/questions/888656/… $\endgroup$ – imranfat Mar 7 '18 at 1:22
33
$\begingroup$

Let $$y=x^2+7x+6$$ $$x=y^2+7y+6$$ thus $$y-x=(x-y)(x+y)+7(x-y)$$ so either

$$x=y$$ or

$$-1=x+y+7$$

Both of these cases reduce the problem to simple quadratic equations.

And you get $$x=-4\pm\sqrt{2},-3\pm\sqrt{3}$$

$\endgroup$
23
$\begingroup$

Define $f(x)\stackrel{\text{def}}{=}x^2+7x+6$. Then, as you've noted, this equation is $$f(f(x))=x\text{.}$$ Certainly any solution of $f(x)=x$ is a solution of your equation, since for such $x$, $f(f(x))=f(x)=x$.

Now here's the trick: since $f(x)-x$ and $f(f(x))-x$ are polynomials such that the roots of the former are roots of the latter, We can use the Division Algorithm to get a new quadratic polynomial $$g(x)=\frac{f(f(x))-x}{f(x)-x}\text{.}$$ Then the roots of $f(f(x))-x$ are the roots of $f(x)-x$ together with the roots of $g(x)$.

$\endgroup$
  • $\begingroup$ You don't even need the trick explicitly: just the observation that solutions to $f(x) = x$ are roots of the original equation give you two roots, then you just have to factor them out (as usual) to find the other two. (Of course it's the same principle, just saying we don't need to explicitly compute the polynomial $g$—though of course it may be easier to do so.) $\endgroup$ – ShreevatsaR Mar 7 '18 at 6:09
  • 1
    $\begingroup$ @ShreevatsaR : true, but such a calculation would involve synthetic division in a quadratic number field—a burden I'm not quite willing to let a precalculus student bear. $\endgroup$ – K B Dave Mar 7 '18 at 6:30

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.