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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?

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  • $\begingroup$ Do you mean $=0$ ? $\endgroup$ Mar 7, 2018 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, 2018 at 1:16
  • $\begingroup$ @idk it says "finding the roots" in the title? $\endgroup$
    – Andrew Li
    Mar 7, 2018 at 1:19
  • $\begingroup$ IDK about it - idk $\endgroup$
    – user535339
    Mar 7, 2018 at 1:21
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    $\begingroup$ This might be interesting: math.stackexchange.com/questions/888656/… $\endgroup$
    – imranfat
    Mar 7, 2018 at 1:22

2 Answers 2

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

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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

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

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  • $\begingroup$ 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.) $\endgroup$ Mar 7, 2018 at 6:09
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    $\begingroup$ @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, 2018 at 6:30

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