Solve the inequality. Write the solution set in interval notation.
-$$( 3 x + 2 ) ( 4 x - 1 ) 0$$
A) $\left[ - \frac { 2 } { 3 } , \frac { 1 } { 4 } \right]$
B) $\left( - \frac { 2 } { 3 } , \frac { 1 } { 4 } \right)$
C) $\left( - \infty , - \frac { 2 } { 3 } \right) \cup \left( \frac { 1 } { 4 } , \infty \right)$
D) $\left( - \infty , - \frac { 2 } { 3 } \right] \cup \left[ \frac { 1 } { 4 } , \infty \right)$

Question

Solve the inequality. Write the solution set in interval notation.
-$$( 3 x + 2 ) ( 4 x - 1 ) > 0$$
A) $\left[ - \frac { 2 } { 3 } , \frac { 1 } { 4 } \right]$
B) $\left( - \frac { 2 } { 3 } , \frac { 1 } { 4 } \right)$
C) $\left( - \infty , - \frac { 2 } { 3 } \right) \cup \left( \frac { 1 } { 4 } , \infty \right)$
D) $\left( - \infty , - \frac { 2 } { 3 } \right] \cup \left[ \frac { 1 } { 4 } , \infty \right)$

Quizplus · Accepted Answer

The inequality $(3x + 2)(4x - 1) > 0$ is satisfied when either both factors are positive or both are negative. Solving for when each factor equals zero gives $x = -\frac{2}{3}$ and $x = \frac{1}{4}$. The inequality is positive outside these points, leading to the solution $\left( -\infty, -\frac{2}{3} \right) \cup \left( \frac{1}{4}, \infty \right)$.

Solve the Inequality (3x+2)(4x−1)>0( 3 x + 2 ) ( 4 x - 1 ) > 0(3x+2)(4x−1)>0

Solve the Inequality $( 3 x + 2 ) ( 4 x - 1 ) > 0$