Find the range of values of x which satisfies the inequality 12x2 < x + 1

A.

-\(\frac{1}{4}\) < x < \(\frac{1}{3}\)

B.

\(\frac{1}{4}\) < x < -\(\frac{1}{3}\)

C.

-\(\frac{1}{3}\) < x < \(\frac{1}{4}\)

D.

-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)

Correct answer is A

12x2 < x + 1

12 - x - 1 < 0

12x2 - 4x + 3x - 1 < 0

4x(3x - 1) + (3x - 1) < 0

Case 1 (+, -)

4x + 1 > 0, 3x - 1 < 0

x > -\(\frac{1}{4}\)x < \(\frac{1}{3}\)

Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0

-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)