Problem 23 - Algebra for College Students

Question

Solve each inequality. $$x^{2}-11 x+28>0$$

Step-by-Step Solution

Answer

The solution is \(x < 4\) or \(x > 7\).

Step 1: Factor the Quadratic Expression

The given inequality is \(x^2 - 11x + 28 > 0\). First, we need to factor the quadratic expression. We need two numbers that multiply to 28 and add up to -11. These numbers are -4 and -7. Therefore, the inequality factors as \((x-4)(x-7) > 0\).

Step 2: Find the Critical Points

The critical points are found by setting each factor equal to zero: \(x-4 = 0\) and \(x-7 = 0\). Solving these gives the critical points \(x = 4\) and \(x = 7\). These points divide the number line into intervals.

Step 3: Test Intervals on the Number Line

The critical points divide the number line into the intervals \((-\infty, 4)\), \((4, 7)\), and \((7, \infty)\). Choose a test point in each interval and determine if the product \((x-4)(x-7)\) is positive in that interval. For \((-\infty, 4)\), use \(x = 0\): \((0-4)(0-7) = 28 > 0\). For \((4, 7)\), use \(x = 5\): \((5-4)(5-7) = -2 < 0\). For \((7, \infty)\), use \(x = 8\): \((8-4)(8-7) = 4 > 0\).

Step 4: Determine the Solution

Based on our tests, the expression \((x-4)(x-7) > 0\) is true for the intervals \((-\infty, 4)\) and \((7, \infty)\). Thus, the solution to the inequality is \(x < 4\) or \(x > 7\).

Key Concepts

Factoring Quadratics

Factoring quadratics entails decomposing a quadratic expression into a product of two binomials. This is often done to simplify solving equations or inequalities involving quadratic expressions. Consider the quadratic inequality \(x^2 - 11x + 28 > 0\).
To factor, you need two numbers that multiply to the constant term, 28, and add up to the linear coefficient, -11. Those numbers are -4 and -7. Thus, the expression \(x^2 - 11x + 28\) factors to \((x-4)(x-7)\).

• The multiplication checks out: \((-4) \times (-7) = 28\).
• The addition is correct: \((-4) + (-7) = -11\).

Factoring makes it easier to analyze the inequality as it breaks down the quadratic into simpler linear terms.

Critical Points

Critical points are the values of \(x\) that make each factor of a factored quadratic equal to zero. They are crucial for understanding where the inequality changes sign.
For the inequality \((x-4)(x-7) > 0\), we find the critical points by solving \(x-4 = 0\) and \(x-7 = 0\). These yield critical points at \(x = 4\) and \(x = 7\).

• To solve \(x-4 = 0\), add 4 to both sides, giving \(x = 4\).
• To solve \(x-7 = 0\), add 7 to both sides, giving \(x = 7\).

These points divide the number line into distinct intervals, which helps in understanding where the inequality holds true.

Testing Intervals

Testing intervals is a method where you assess the sign of the product of factors in various sections between critical points. This establishes where the inequality holds.
The critical points \(x = 4\) and \(x = 7\) divide the number line into three intervals: \((-fty, 4)\), \((4, 7)\), and \((7, fty)\). For each interval, select a test point:

• In \((-fty, 4)\), use \(x = 0\). Calculate \((0-4)(0-7) = 28 > 0\).
• In \((4, 7)\), use \(x = 5\). Calculate \((5-4)(5-7) = -2 < 0\).
• In \((7, fty)\), use \(x = 8\). Calculate \( (8-4)(8-7) = 4 > 0 \).

This process checks where the product is positive, indicating the interval's validity for the inequality.

Inequality Solutions

Solving quadratic inequalities involves determining which sections of the number line satisfy the inequality’s condition. After testing the intervals, we need to state the solution for \((x-4)(x-7) > 0\).
From our interval testing, the inequality holds true in \((-fty, 4)\) where the product is positive and in \((7, fty)\) where the product is again positive.

• In \((4, 7)\), the product is negative, so the inequality does not hold.

Thus, the solution to the inequality is \(x < 4\) or \(x > 7\).
This means any number less than 4 or greater than 7 will satisfy the original inequality \(x^2 - 11x + 28 > 0\).