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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?

Blue
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2 Answers2

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

Rene Schipperus
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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

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

K B Dave
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  • 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.) – ShreevatsaR Mar 07 '18 at 06:09
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    @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. – K B Dave Mar 07 '18 at 06:30