Problem 39 - Precalculus Student Solutions Manual 5th

Question

Solve each quadratic inequality. Write each solution set in interval notation. $$x^{2}-x-6>0$$

Step-by-Step Solution

Answer

\((-\infty, -2) \cup (3, \infty)\)

Step 1 - Factor the quadratic expression

Rewrite the quadratic expression in factored form. Find the values of x that satisfy the equation. \(x^2 - x - 6 = (x - 3)(x + 2)\) By setting the factors to 0, we find the roots of the equation: \(x - 3 = 0 \ x = 3\)\(x + 2 = 0 \ x = -2\)

Step 2 - Determine intervals on the real number line

Use the roots to divide the number line into intervals. The intervals are: \((-\infty, -2)\), \((-2, 3)\), and \((3, \infty)\).

Step 3 - Test points in each interval

Choose a test point from each interval and determine the sign of the expression \((x - 3)(x + 2)\) in these intervals: For \((-\infty, -2)\), use x = -3: \((-3 - 3)(-3 + 2) = -6(-1) = 6 > 0\) For \((-2, 3)\), use x = 0: \((0 - 3)(0 + 2) = -3(2) = -6 < 0\) For \((3, \infty)\), use x = 4: \((4 - 3)(4 + 2) = 1(6) = 6 > 0\)

Step 4 - Write the solution set

The expression is positive in the intervals \((-\infty, -2)\) and \((3, \infty)\). Thus, the solution set in interval notation is: \((-\infty, -2) \cup (3, \infty)\)

Key Concepts

Factoring Quadratics

To solve quadratic inequalities, the first step is often to factor the quadratic expression. Factoring simplifies expressions and helps in finding the roots. Let's take the quadratic expression from our problem: \(x^2 - x - 6\).The goal is to rewrite it as a product of two binomials. The factors of \(-6\) that add up to \(-1\) are \(3\) and \(-2\). So, we can factor the expression as:\((x - 3)(x + 2)\).Factoring quadratic expressions makes it easier to identify the points where the expression crosses the x-axis (the roots). Setting each factor equal to zero gives us the roots of the inequality: \[(x - 3) = 0 \rightarrow x = 3\]\[(x + 2) = 0 \rightarrow x = -2\].

Interval Notation

Once we have the roots, we can use them to divide the number line into intervals. These intervals are crucial because they help us determine where the quadratic expression is positive or negative. From our problem, the roots are \(x = 3\) and \(x = -2\). So, the number line is divided into three intervals:

• \((-\text{infty}, -2)\)
• \((-2, 3)\)
• \((3, \text{infty})\)

These intervals represent sections of the number line that will be tested to see where the original inequality holds true. Interval notation provides a concise way to represent these ranges.

Sign Testing

Sign testing involves selecting test points from each interval to determine whether the quadratic expression is positive or negative within those intervals. This helps us identify which intervals satisfy the inequality. For our three intervals:

• In \((-\text{infty}, -2)\), let's use \(x = -3\). Substituting, we get: \(( -3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0\)
• In \((-2, 3)\), let's use \(x = 0\). Substituting, we get: \((0 - 3)(0 + 2) = (-3)(2) = -6 < 0\)
• In \((3, \text{infty})\), let's use \(x = 4\). Substituting, we get: \((4 - 3)(4 + 2) = (1)(6) = 6 > 0\)

By testing these points, we see that the expression is positive in the intervals \((-\text{infty}, -2)\) and \((3, \text{infty})\). Finally, we can write the solution set in interval notation as: \((-\text{infty}, -2) \text{cup} (3, \text{infty})\). Sign testing is a powerful tool for precisely determining where quadratic expressions meet conditions imposed by inequalities.