If you mean $p_-$ or $p_+$, then indeed, they are the roots of $p$, so $p(p_+) = p(p_-) = 0$, but they are not necessarily the roots of $p(f(x))$.įor example, let $p(x) = x^2-16x+64 = (x-8)^2$ and $f(x) = x^3$. Not sure how a function can be a root of itself. These final solutions $x^*$ will be solutions of (1) and roots of $p(f(x))$. To do that, as you correctly note, you first find the roots of $p(x)$, say $p_+$ and $p_-$, so that $p(p_+) = 0 = p(p_-)$, and then solve $f(x) = p_+$ and $f(x) = p_-$, obtaining some solutions, say $x^*$. In other words, x is a root of the quadratic equation f(x), if f().
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