Problem 6 - Algebra for College Students

Question

Solve each inequality and graph its solution set on a number line. $$(3 x+2)(2 x-3) \geq 0$$

Step-by-Step Solution

Answer

The solution is \([-\frac{2}{3}, \frac{3}{2}] \cup (\frac{3}{2}, \infty)\).

Step 1: Set Each Factor to Zero

We first solve the inequalities where each of the factors would equal zero to find the critical points: \(3x + 2 = 0\) and \(2x - 3 = 0\). This will help us find where the expression changes its sign.

Step 2: Solve for Critical Points

Solve the equation \(3x + 2 = 0\) to get \(x = -\frac{2}{3}\). Solve \(2x - 3 = 0\) to get \(x = \frac{3}{2}\). These values divide the number line into intervals for testing.

Step 3: Determine Intervals

The critical points \(x = -\frac{2}{3}\) and \(x = \frac{3}{2}\) divide the number line into three intervals: \((-\infty, -\frac{2}{3})\), \((-\frac{2}{3}, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\). We will pick sample points in each interval to determine if each interval satisfies the inequality \((3x+2)(2x-3) \geq 0\).

Step 4: Test Intervals with Sample Points

Choose a test point in each interval: for \((-\infty, -\frac{2}{3})\), choose \(x = -1\); for \((-\frac{2}{3}, \frac{3}{2})\), choose \(x = 0\); and for \((\frac{3}{2}, \infty)\), choose \(x = 2\).- For \(x = -1\): \((3(-1) + 2)(2(-1) - 3) = (-1)(-5) = 5\) so this interval satisfies the inequality.- For \(x = 0\): \((3(0) + 2)(2(0) - 3) = (2)(-3) = -6\) so this interval does not satisfy the inequality.- For \(x = 2\): \((3(2) + 2)(2(2) - 3) = (8)(1) = 8\) so this interval satisfies the inequality.

Step 5: Include Critical Points

Evaluate the original inequality at the critical points. For \(x = -\frac{2}{3}\) and \(x = \frac{3}{2}\), the factor expression becomes zero and thus satisfies \((3x+2)(2x-3) \geq 0\). Hence, include both critical points in the solution.

Step 6: Write Final Solution and Graph

Combine the satisfying intervals and include the critical points: the solution is \([-\frac{2}{3}, \frac{3}{2}]\) and \((\frac{3}{2}, \infty)\). On a number line, represent this by shading the area from \(-\frac{2}{3}\) to \frac{3}{2}, including closed circles at these points, and shading the area from \frac{3}{2}\ to positive infinity with an open circle at \(\frac{3}{2}\).

Key Concepts

Critical Points

When we solve inequalities like \((3x+2)(2x-3) \geq 0\), it's crucial to identify critical points. These points occur where each factor equals zero. By setting \(3x + 2 = 0\) and \(2x - 3 = 0\), we determine the values of \(x\) where the expression can change its sign. Solving these equations gives us the critical points: \(x = -\frac{2}{3}\) and \(x = \frac{3}{2}\).
The importance of critical points lies in their role as boundaries. They are the points at which the sign of the product may switch from positive to negative (or vice versa). Without identifying these, we cannot accurately determine where the inequality holds true.
After finding these critical points, they guide us in dividing the number line into intervals for further analysis.

Number Line Graph

A number line graph is a visual representation of the solutions to an inequality. Once we have our critical points, we can mark them on a number line. This division creates segments that help evaluate where the inequality holds true.
For our problem, marking \(x = -\frac{2}{3}\) and \(x = \frac{3}{2}\) on the number line breaks it into three specific intervals: \((-\infty, -\frac{2}{3})\), \((-\frac{2}{3}, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\).
In this process, the number line graph allows us to see at a glance where the solutions might lie. Based on our interval testing results, we know which parts of the number line to include in the solution set. Closed circles indicate that the critical points are part of the solution, meaning the inequality holds true at these points, while open circles suggest they are not included.

Factor Expression

In inequalities involving factor expressions, such as \((3x+2)(2x-3)\), understanding each part is essential. Each factor can potentially change the sign of the expression's product.
To analyze a factor expression:

• Set each factor to zero: \(3x + 2 = 0\) and \(2x - 3 = 0\)
• Find solutions to identify critical points, which in this case led us to \(x = -\frac{2}{3}\) and \(x = \frac{3}{2}\)

The factor expression helps determine where the signs of intervals switch. Recognizing how these factors influence the inequality lets us precisely determine the solutions. A positive product means both factors are either positive or negative, while a negative product indicates one positive and one negative.

Interval Testing

Interval testing is a technique to determine which intervals satisfy the inequality. Once the critical points divide the number line into intervals, we select test points from each one.
Here's the plan:

• Choose a test point in\((-\infty, -\frac{2}{3})\), say \(x = -1\)
• Choose \(x = 0\) for \((-\frac{2}{3}, \frac{3}{2})\)
• Choose \(x = 2\) for \((\frac{3}{2}, \infty)\)

Using these test points, substitute into the inequality and check the results.For instance, \((3(-1) + 2)(2(-1) - 3) = 5\) shows the first interval works, while others may not. Confirm which intervals make the expression \(\geq 0\) to form the solution set. Only include intervals that satisfy the inequality.