Question

Solve for x, (3x – 1)(2x + 3) < 0.​

Answer 1

Solution:

Given Inequality:

$\rm \longrightarrow (3x - 1)(2x + 3) < 0$

We have to know the basic concepts before solving this problem.

We know that the product of two quantities is negative if any one of the quantities is negative.

Here, the product of (3x - 1) and (2x + 3) is less than 0 means either (3x - 1) is negative or (2x + 3)

So, two cases arises here.

$\rm \longrightarrow Case \: 1 = \begin{cases} \rm 3x - 1 < 0 \\ \rm 2x + 3 > 0 \end{cases}$

$\rm \longrightarrow Case \: 2= \begin{cases} \rm 3x - 1 > 0 \\ \rm 2x + 3<0 \end{cases}$

Solving (i), we get:

$\rm \longrightarrow 3x - 1 < 0$

$\rm \longrightarrow x< \dfrac{1}{3}$

$\rm \longrightarrow 2x+3 > 0$

$\rm \longrightarrow x> \dfrac{ - 3}{2}$

Combining both, we get:

$\rm \longrightarrow \dfrac{ - 3}{2} < x < \dfrac{1}{3}$

$\rm \longrightarrow x \in\bigg(\dfrac{ - 3}{2}, \dfrac{1}{3} \bigg)$

Solving (ii), we get:

$\rm \longrightarrow 3x - 1 > 0$

$\rm \longrightarrow x > \dfrac{1}{3}$

$\rm \longrightarrow 2x+3 < 0$

$\rm \longrightarrow x < \dfrac{ - 3}{2}$

Combining both, we get:

$\rm \longrightarrow x > \dfrac{1}{3} \: and \: x < \dfrac{ - 3}{2}$

Which is not possible at the same time. Therefore:

$\rm \longrightarrow x \in \phi$

Combining both solutions, we get:

$\rm \longrightarrow x \in\bigg(\dfrac{ - 3}{2}, \dfrac{1}{3} \bigg) \: \cup \: \phi$

$\rm \longrightarrow x \in\bigg(\dfrac{ - 3}{2}, \dfrac{1}{3} \bigg)$

★ Which is our required answer.

Answer:

$\rm \hookrightarrow x \in\bigg(\dfrac{ - 3}{2}, \dfrac{1}{3} \bigg)$