Problem 29 - Precalculus : Building Concepts and Connections

Question

Solve the rational inequality. $$\frac{x^{2}-4}{x-3} \leq 0$$

Step-by-Step Solution

Answer

The solution to the inequality \(\frac{x^{2}-4}{x-3} \leq 0\) is \(x ∈ (-∞, -2] ∪ (2, 3]\)

Step 1: Solve the quadratic equation

Firstly, factorize the quadratic equation in the numerator \(x^{2}-4\). It can be factorized as \((x-2)(x+2)\). So, the equation becomes \(\frac{(x-2)(x+2)}{x-3} \leq 0\).

Step 2: Identify the critical numbers

The critical numbers are the solutions of the equation \(\frac{(x-2)(x+2)}{x-3} = 0\). Equating both numerator and denominator to zero gives the critical numbers as \(x = 2, -2, 3\).

Step 3: Analyze the intervals defined by critical numbers

The critical numbers \(2, -2, 3\) divide the number line into four intervals: \((-∞, -2), (-2, 2), (2, 3)\), and \( (3, ∞)\). Pick a test number from each interval, substitute it into \(\frac{(x-2)(x+2)}{x-3}\), and determine whether the result is positive or negative.

Step 4: Solution of the inequality

The solutions of the inequality \(\frac{(x-2)(x+2)}{x-3} \leq 0\) are the union of the intervals where the expression is negative or zero. In this case, only in the intervals \((-∞, -2]\) and \((2, 3]\), the inequality is less than or equal to zero, and hence the solution is \(x ∈ (-∞, -2] ∪ (2, 3]\).

Key Concepts

Quadratic Factorization

Quadratic factorization is a crucial step when solving rational inequalities like \(\frac{x^{2}-4}{x-3} \leq 0\). When you encounter a quadratic expression, such as \(x^2 - 4\), breaking it down into its factors is essential. Quadratics of the form \(x^2 - a^2\) are often best tackled using the difference of squares formula:

• \((a^2 - b^2) = (a-b)(a+b)\)

By applying this, we recognize that \(x^2 - 4\) is identical to the expression \((x - 2)(x + 2)\).
This form is much simpler to work with, especially when integrated into a rational inequality.
Using factorization not only helps solve the inequality but also gives insight into the behavior of the function, enabling the identification of critical points and intervals easily.

Critical Numbers

Critical numbers are pivotal in solving inequalities as they mark where the expression is equal to zero or undefined. For the expression \(\frac{(x-2)(x+2)}{x-3}\), the critical numbers are found by setting each factor equal to zero or by detecting division by zero:

• The numerator, \((x-2)(x+2)\), equals zero when \(x = 2\) or \(x = -2\).
• The denominator creates a division error when \(x=3\), so it’s a critical point too.

These values, \(x = -2, 2, 3\), indicate where the expression can change its sign, creating boundaries for interval testing.
Understanding critical numbers in this way allows you to map out where an inequality could potentially be positive, negative, or zero.

Interval Testing

Interval testing is used to determine which intervals satisfy a rational inequality. The critical points from \(\frac{(x-2)(x+2)}{x-3} \leq 0\), \(-2, 2,\) and \(3\), divide the number line into four distinct intervals:

• \((-∞, -2)\)
• \((-2, 2)\)
• \((2, 3)\)
• \((3, ∞)\)

By selecting a test point within each interval, you can substitute it back into the inequality to check if it results in a positive or negative outcome. For example:

• Select \(-3\) for the interval \((-∞, -2)\), resulting in a negative expression.
• Choose \(0\) in \((-2, 2)\), and the expression resolves as positive.
• Try \(2.5\) in \((2, 3)\), and you'll find negative results once more.
• Pick \(4\) for \((3, ∞)\), resulting in positive.

The intervals where the expression is zero or negative are combined to provide the solution: \((-∞, -2] \cup (2, 3]\).
Interval testing connects critical points back to the inequality, ensuring solutions are exact and informative for solving inequalities such as this.