# 1963 AHSME Problems/Problem 28

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## Problem

Given the equation $3x^2 - 4x + k = 0$ with real roots. The value of $k$ for which the product of the roots of the equation is a maximum is:

$\textbf{(A)}\ \frac{16}{9}\qquad \textbf{(B)}\ \frac{16}{3}\qquad \textbf{(C)}\ \frac{4}{9}\qquad \textbf{(D)}\ \frac{4}{3}\qquad \textbf{(E)}\ -\frac{4}{3}$

## Solution

By Vieta's Formulas, the product of the roots is $\frac{k}{3}$. This value increases as $k$ increases.

Also, the quadratic’s roots are real, then the discriminant is greater than or equal to zero, so $$16-12k \ge 0$$ $$-12k \ge -16$$ $$k \le \frac{4}{3}$$

Thus, the $k$ value that maximizes the product of the roots is $\frac{4}{3}$, which is answer choice $\boxed{\textbf{(D)}}$.

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