Finding the roots of $(x^2+7x+6)^2 + 7(x^2+7x+6) + 6=x$

Question

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?

Answer 1

Define $f(x)\stackrel{\text{def}}{=}x^2+7x+6$. Then, as you've noted, this equation equivalent to finding the roots to the quartic $$h(x) = f(f(x))-x\text{.}$$

Write $f[x,y] = x + y + 7$ for the divided difference of $f$, i.e., the unique bivariate polynomial such that $$f(x) - f(y) = (x-y) f[x,y]\text{,}$$ as calculated via the Division Algorithm.

Then the trick is to add and subtract $f(x)$ and recognize the divided difference:

$$\begin{split}h(x) &= f(f(x))-x \\ &= f(f(x)) - f(x) + f(x) - x \\ &= (f(x) - x)f[f(x),x] + f(x) - x \\ &= k(x)g(x) \end{split}$$

where $k(x)$ and $g(x)$ are quadratic polynomials

$$\begin{align} k(x)&\stackrel{\text{def}}{=}f(x) - x &&= x^2 + 6x + 6 &&= (x+3)^2 - 3 \\ g(x)&\stackrel{\text{def}}{=}f[f(x),x] +1 &&= x^2 + 8x + 14 &&= (x+4)^2 - 2 \text{.} \end{align}$$

Answer 2

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