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

1971 AMC 12 Problem 8

Attempt Problem 8 of the 1971 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1971 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

The solution set of

6x^2+5x\lt4

is the set of all values of

x

such that:

-2\lt x\lt1

-\dfrac43\lt x\lt\dfrac12

-\dfrac12\lt x\lt\dfrac43

x\lt\dfrac12

x\gt-\dfrac43

x\lt-\dfrac43

x\gt\dfrac12

Move

4

to the left and factor

The product

(3x+4)(2x-1)

is negative between its roots

The inequality is

\begin{aligned} 6x^2+5x-4&\lt0,\\ (3x+4)(2x-1)&\lt0. \end{aligned}

The roots are

-\frac{4}{3}

and

\frac{1}{2},

and the upward-opening quadratic is negative between them.

Therefore, the correct answer is B.