Problem 38 - Algebra for College Students

Question

Solve each inequality. $$2 x^{2}-18 \geq 0$$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -3] \cup [3, \infty)\).

Step 1: Identify the Expression to Solve

We need to solve the inequality \(2x^2 - 18 \geq 0\). This is a quadratic inequality.

Step 2: Simplify the Inequality

Divide every term by 2 to simplify: \(x^2 - 9 \geq 0\). This simplifies the inequality to a more manageable form.

Step 3: Factoring the Quadratic Expression

Recognize that \(x^2 - 9\) can be factored as the difference of squares: \((x + 3)(x - 3) \geq 0\).

Step 4: Determine Critical Points

Set each factor equal to zero to find the critical points: \(x + 3 = 0\) gives \(x = -3\) and \(x - 3 = 0\) gives \(x = 3\). These points divide the real number line into intervals.

Step 5: Test the Intervals

Test intervals between the critical points. The intervals are: \((-\infty, -3)\), \([-3, 3]\), and \((3, \infty)\). Choose a test point from each interval to see where the inequality holds:1. Test point \(x = -4\) in \((-\infty, -3)\): \((-4 + 3)(-4 - 3) = 1\times (-7) < 0\).2. Test point \(x = 0\) in \([-3, 3]\): \((0 + 3)(0 - 3) = 3\times (-3) < 0\).3. Test point \(x = 4\) in \((3, \infty)\): \((4 + 3)(4 - 3) = 7\times 1 > 0\).

Step 6: Include Critical Points

Since the original inequality is \( (x+3)(x-3) \geq 0\), include the points where it equals 0. The inequality is satisfied at \(x = -3\) and \(x = 3\).

Step 7: State the Solution Set

Combine the results of the interval tests and critical points. The solution set is \((-\infty, -3] \cup [3, \infty)\).

Key Concepts

Factoring Quadratic Expressions

When solving quadratic inequalities, a great first step is factoring the quadratic expression. Factoring involves finding two numbers or expressions that multiply together to recreate the original quadratic expression. In this case, we have the expression \( x^2 - 9 \). This particular expression can be rewritten using the difference of squares method.
The difference of squares is a pattern that can be identified in the form \( a^2 - b^2 \), which can be factored into \( (a + b)(a - b) \). For \( x^2 - 9 \), recognize that this equals \( x^2 - 3^2 \). Therefore, it factors as \( (x + 3)(x - 3) \).
Factoring is useful because it helps decompose the quadratic into simpler linear factors, making it easier to analyze or solve any given inequality.

Critical Points in Inequalities

Critical points are values that make each factor of the quadratic expression equal to zero. These points are important when analyzing where a quadratic inequality holds true. Once the quadratic expression has been factored, you can find the critical points by setting each factor equal to zero.
For the expression \((x + 3)(x - 3)\), setting each factor to zero, we have:

• \( x + 3 = 0 \), which gives us \( x = -3 \)
• \( x - 3 = 0 \), which gives us \( x = 3 \)

These critical points, \( x = -3 \) and \( x = 3 \), divide the number line into regions or intervals. Identifying critical points helps in testing intervals to determine where the inequality is true.

Interval Testing in Inequalities

Interval testing is a method to determine where a quadratic inequality holds true by selecting test points from the intervals created by critical points.
After finding the critical points, the real number line is divided into segments:

• \( (-\infty, -3) \)
• \( [-3, 3] \)
• \( (3, \infty) \)

Choose test points from each interval and substitute them into the factored inequality \( (x + 3)(x - 3) \geq 0 \). For example:

• Test point \( x = -4 \) in \( (-\infty, -3) \) results in a negative product, indicating \( (x + 3)(x - 3) < 0 \)
• Test point \( x = 0 \) in \( [-3, 3] \) also results in a negative product
• Test point \( x = 4 \) in \( (3, \infty) \) results in a positive product, meaning \( (x + 3)(x - 3) > 0 \)

The original inequality \( (x + 3)(x - 3) \geq 0 \) includes points where the expression equals zero, so \( x = -3 \) and \( x = 3 \) are part of the solution. Combining these findings leads to the conclusion that the inequality holds true for: \( (-\infty, -3] \cup [3, \infty) \).