Finding the Roots of a Certain Quartic Equation

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?

Solution 1:

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

Solution 2:

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 $g(x)$.