# Roots of Equation

Let the roots of the equation $$x^3 + px^2 + qx + r = 0$$ be in arithmetic progression. Show that $$p^2 \ge 3q.$$

Attempt: Let the roots be $\alpha$, $\beta$, and $\gamma$. Then $$\sum\alpha=-p, \quad \sum\alpha\beta=q, \quad\text{and}\quad \alpha\beta \gamma=-r.$$ Since roots are in $AP$, we have $2\beta=\alpha+\gamma$.

WLOG the roots are $a-d,a,a+d$
So, $-p=\cdots=3a$
and $q= a(a+d)+(a+d)(a-d)+(a-d)a=3a^2-d^2$
You can shift the unknown $x$ by $-p/3$ to cancel the quadratic term and get a polynomial with $p'=0,q'=q-p^2/3$, while the roots still form an AP.
Then from $p'=0$, the roots must be $-t,0,t$, and by Vieta, $q'=-t^2/3<0$. This implies $q-p^3/3<0$.