1971 AHSME Problems/Problem 8


The solution set of $6x^2+5x<4$ is the set of all values of $x$ such that

$\textbf{(A) }\textstyle -2<x<1\qquad \textbf{(B) }-\frac{4}{3}<x<\frac{1}{2}\qquad \textbf{(C) }-\frac{1}{2}<x<\frac{4}{3}\qquad \\ \textbf{(D) }x<\textstyle\frac{1}{2}\text{ or }x>-\frac{4}{3}\qquad \textbf{(E) }x<-\frac{4}{3}\text{ or }x>\frac{1}{2}$


We are solving the inequality $6x^2 + 5x - 4  < 0.$ This can be factored as \[(2x-1)(3x+4) < 0\]

The graph of this inequality is a parabola facing upwards, so the area between the roots satisfies the equation. The solution is $x \in [-\frac{4}{3}, \frac{1}{2}]$ and the answer is $\textbf{(C)}.$

-edited by coolmath34