Problem 35 - Algebra for College Students

Question

Solve each inequality. $$(x+1)(x-3)^{2}>0$$

Step-by-Step Solution

Answer

The solution is \((-1, 3) \cup (3, \infty)\).

Step 1: Identify critical points

First, find the critical points where the inequality changes sign. These occur where each factor equals zero: \(x + 1 = 0\) gives \(x = -1\), and \((x - 3)^2 = 0\) gives \(x = 3\). Thus, the critical points are \(x = -1\) and \(x = 3\).

Step 2: Determine intervals to test

Using the critical points, divide the number line into intervals: \((-\infty, -1)\), \((-1, 3)\), and \((3, \infty)\). We will test these intervals to determine where the inequality \((x+1)(x-3)^2 > 0\) holds.

Step 3: Test each interval

Test the sign of \((x + 1)(x - 3)^2\) in each interval:1. Choose \(x = -2\) in \((-\infty, -1)\): \((x + 1)(x - 3)^2 = (-2 + 1)((-2 - 3)^2) = (-1)(25) < 0\).2. Choose \(x = 0\) in \((-1, 3)\): \((x + 1)(x - 3)^2 = (0 + 1)((0 - 3)^2) = 1(9) > 0\).3. Choose \(x = 4\) in \((3, \infty)\): \((x + 1)(x - 3)^2 = (4 + 1)((4 - 3)^2) = 5(1) > 0\).

Step 4: Determine solution set

Collect the intervals where the test was positive: \((-1, 3)\) and \((3, \infty)\). Since \((x - 3)^2\) is always positive or zero and only \(x + 1\) affects the sign, the factor assay means \(x = 3\) does not make the product positive (only zero), hence is not included in the solution set.

Key Concepts

Critical Points

Finding the critical points is an essential step in solving inequalities like \((x+1)(x-3)^{2} > 0\). Critical points are those values of \(x\) that make each factor in the inequality equal to zero. These points are crucial as they divide the number line into intervals where the inequality might change its sign. For the given inequality, set each factor to zero:

• \(x + 1 = 0\) results in \(x = -1\)
• \((x - 3)^2 = 0\) provides \(x = 3\)

Thus, the critical points are \(x = -1\) and \(x = 3\). Once these points are identified, they help in determining the intervals to test to solve the inequality further. It's important to note that these points might or might not be included in the solution set, depending on whether or not they satisfy the inequality. In this case, they merely help us divide the number line.

Interval Testing

Interval testing follows naturally after identifying critical points. This process helps to determine in which intervals the inequality holds true. By using the critical points we found, we can divide the number line into specific intervals. For \((x+1)(x-3)^{2} > 0\), the intervals are:

• \((-∞, -1)\)
• \((-1, 3)\)
• \((3, ∞)\)

Each of these intervals needs to be tested by choosing a test point within it. The sign of the product will tell us if that interval makes the inequality true or false. For instance:- Within \((-∞, -1)\), choosing \(x = -2\), we find the expression negative.- For \((-1, 3)\), \(x = 0\) gives a positive result, satisfying the inequality.- And \((3, ∞)\) with \(x = 4\) also results in a positive expression.These tests lead to identifying the intervals that satisfy the given inequality, helping us pinpoint the solution set accurately.

Quadratic Inequalities

Quadratic inequalities involve expressions that include a quadratic component, giving them a unique behaviour at critical points and intervals between them. Inequalities like \((x+1)(x-3)^2 > 0\) typically consist of several factors multiplied together. This structure means that the sign of each interval is often determined by the factors' behaviour over the entire range.

Understanding the role of the quadratic term \((x - 3)^2\) is key here. Since squares are always non-negative, \((x - 3)^2\) is always \(≥ 0\). Thus, it does not change the sign of the inequality, only \(x+1\) influences when the product becomes positive or zero.

The inequality > 0 means we are looking for where the entire product is greater than zero, excluding where it is exactly zero. Therefore, assessing each factor's sign helps understand intervals where the product would yield positive values. By analyzing these relationships and testing them through chosen interval points, you can accurately determine what values of \(x\) satisfy the inequality.