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Let $\alpha$ be the only real root of $p(x)=x^3-3x^2+5x-17$ and $\beta$ the only real root of $q(x)=x^3-3x^2+5x+11$. Compute $\alpha + \beta$.

I've noticed that the two graphs are just the same shifted vertically (the first coefficients are equal except for the last), but I don't know if it is useful.

(This is supposed to be doable without any calculator)

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Re-arrange $p(x)$ and $q(x)$:

\begin{align*} p(x) &= x^3-3x^2+5x-17 \\ &= (x-1)^3+2(x-1)-14 \\ q(x) &= x^3-3x^2+5x+11 \\ &= (x-1)^3+2(x-1)+14 \end{align*}

Note that $$p(1+x) = -q(1-x)$$

If $p(1+x)=-q(1-x)=0$, then $$ \left \{ \begin{align*} 1+x &= \alpha \\ 1-x &= \beta \end{align*} \right.$$

$$\fbox{$\alpha+\beta=2$}$$