Let the equation $x^3 - 16x^2 + px^2 - 256x + q = 0$ has four positive real roots in G.P., then $p + q$ is equal to
Step-by-Step Solution
Key Concept: For roots in G.P., the product of extremes equals the product of means, which simplifies Vieta relations.
Given the quartic $x^4 - 16x^3 + px^2 - 250x + q = 0$ with roots in G.P., let roots be $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$. Using Vieta's formulas: $x_1 x_4 = x_2 x_3$ gives $a^2 = x_2 x_3$, and $(x_1 + x_4) + (x_2 + x_3) = 16$ constrains the sum of paired roots.
Correct Answer: 352