To solve the inequality $(x+3)(2x-3) > 0$ using the case method, we will follow these steps:

Step 1: Find the critical points

The critical points are the values of $x$ where the expression $(x+3)(2x-3)$ equals 0. So, solve:

$x+3 = 0 \quad \Rightarrow \quad x = -3$

$2x - 3 = 0 \quad \Rightarrow \quad x = \frac{3}{2}$

Thus, the critical points are $x = -3$ and $x = \frac{3}{2}$.

Step 2: Set up intervals based on critical points

The real number line is divided into three intervals by these critical points:

• $(-\infty, -3)$
• $(-3, \frac{3}{2})$
• $(\frac{3}{2}, \infty)$

Now, we will test the sign of the expression $(x+3)(2x-3)$ in each of these intervals.

Step 3: Test each interval

We will choose test points in each interval and substitute them into the expression $(x+3)(2x-3)$ to determine its sign.

1. Interval $(-\infty, -3)$: Choose a test point, say $x = -4$: $(x+3)(2x-3) = (-4+3)(2(-4)-3) = (-1)(-8-3) = (-1)(-11) = 11 \quad \text{(positive)}$

2. Interval $(-3, \frac{3}{2})$: Choose a test point, say $x = 0$: $(x+3)(2x-3) = (0+3)(2(0)-3) = 3(-3) = -9 \quad \text{(negative)}$

3. Interval $(\frac{3}{2}, \infty)$: Choose a test point, say $x = 2$: $(x+3)(2x-3) = (2+3)(2(2)-3) = 5(4-3) = 5 \quad \text{(positive)}$

Step 4: Analyze the solution

We are looking for where the expression $(x+3)(2x-3) > 0$. From our tests:

• In the interval $(-\infty, -3)$, the expression is positive.
• In the interval $(-3, \frac{3}{2})$, the expression is negative.
• In the interval $(\frac{3}{2}, \infty)$, the expression is positive.

Thus, the inequality $(x+3)(2x-3) > 0$ holds in the intervals $(-\infty, -3)$ and $(\frac{3}{2}, \infty)$.

Final Solution:

The solution to the inequality is: $x \in (-\infty, -3) \cup \left(\frac{3}{2}, \infty\right)$

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Would you like a more detailed explanation of any part of this process, or do you have any other questions?

Here are some related questions to explore:

1. How would the solution change if the inequality were $(x+3)(2x-3) \leq 0$?
2. How do we handle quadratic inequalities like this one using the sign chart method?
3. How can we solve a rational inequality using the same method?
4. What does the graph of the function $y = (x+3)(2x-3)$ look like?
5. How do you solve a system of inequalities using the case method?

Tip: Always check if the inequality includes equalities ($\geq$ or $\leq$), which can affect whether critical points are part of the solution set.