Let $\alpha$ and $\beta$ be the roots of the equation $px^{2} + qx + r = 0$. If $p, q, r$ are in AP and $\alpha + \beta = 4$, then $\alpha \beta$ is equal to

Using the sum of roots for px^2 + qx + r = 0, alpha + beta is -q/p, which equals 4, so q = -4p. Since p, q, r are in arithmetic progression, we have 2q = p + r, and substituting q gives 2(-4p) = p + r, meaning r = -9p. The product of the roots, alpha * beta, is r/p, which equals -9p/p = -9.

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