Problem 24 - Algebra 2

Question

Solve each inequality using a graph, a table, or algebraically. $$ x^{2}+2 x \geq 24 $$

Step-by-Step Solution

Answer

The solution set is \( x \in (-\infty, -6] \cup [4, \infty) \).

Step 1: Rearrange the inequality

First, rearrange the inequality to have 0 on one side. Starting with \( x^2 + 2x \geq 24 \), subtract 24 from both sides to get: \( x^2 + 2x - 24 \geq 0 \).

Step 2: Factor the quadratic expression

Attempt to factor the quadratic expression \( x^2 + 2x - 24 = 0 \). Look for two numbers that multiply to -24 and add to 2. These numbers are 6 and -4, resulting in the factorization: \( (x + 6)(x - 4) \geq 0 \).

Step 3: Determine the zeros

Find the zeros by setting each factor equal to zero: \( x + 6 = 0 \) gives \( x = -6 \) and \( x - 4 = 0 \) gives \( x = 4 \). These zeros are the critical points where the expression may change signs.

Step 4: Test intervals around zeros

Divide the number line into intervals based on the zeros: \( (-\infty, -6) \), \( (-6, 4) \), and \( (4, \infty) \). Test each interval by picking a test point: \(-7, 0, 5\) respectively. Substitute each into \( (x+6)(x-4) \).

Step 5: Analyze the sign of each interval

For \( x = -7 \), \( (-7+6)(-7-4) = (-1)(-11) = 11 > 0 \). For \( x = 0 \), \( (0+6)(0-4) = 6(-4) = -24 < 0 \). For \( x = 5 \), \( (5+6)(5-4) = 11 \cdot 1 = 11 > 0 \).

Step 6: Write the solution set

Since the inequality is \( \geq 0 \), include intervals where the product is positive or zero. Therefore, the solution is \( x \in (-\infty, -6] \cup [4, \infty) \).

Key Concepts

Quadratic Equations

Understanding quadratic equations is essential in solving inequalities involving a square term. A quadratic equation is typically in the form of \( ax^2 + bx + c = 0 \). Here, the equation \( x^2 + 2x - 24 \) comes into play after rearranging the original inequality. It's crucial to identify the standard form as it allows for further methods like factoring or using the quadratic formula.
Quadratic equations often appear in parabolic shapes when graphed. They have one or two solutions, known as roots or zeros. To solve them, one can set the equation to zero and use various algebraic methods, like factoring. Recognizing the structure of quadratic equations is a powerful tool for both graphical and algebraic solutions.

Factoring

Factoring is a technique used to simplify quadratic equations, making them easier to solve. The idea is to break down the equation into simpler expressions that can be multiplied together to bring back the original equation. In the case of our equation, \( x^2 + 2x - 24 \) was factored into \( (x + 6)(x - 4) = 0 \).
You look for two numbers that multiply to give the constant term (-24 in this case) and add up to the coefficient of the linear term (2 here). The numbers that satisfy these conditions are 6 and -4.
Factoring transforms the equation into a product of linear factors that are easier to solve. Once factored, we can find the solutions or zeros by setting each factor to zero. This step is crucial for solving inequalities.

Inequality Solutions

Inequality solutions involve identifying the set of values that satisfy an inequality. After factoring and finding zeros, those zeros are critical points that divide the number line into intervals. Determining which intervals meet the inequality's condition involves testing points within those intervals.
For \( (x + 6)(x - 4) \geq 0 \), you identify the intervals \((-\infty, -6)\), \((-6, 4)\), and \((4, \infty)\). By testing a sample point in each interval, such as \(-7, 0,\) and \(5\), we find whether the inequality holds. Only intervals with positive results or zeros are part of the solution set.
Inequalities often lead to ranges of values as solutions, represented with interval notation, such as \(x \in (-\infty, -6] \, \cup \, [4, \infty)\).

Critical Points

Critical points in inequalities are the values where the mathematical expression may change its sign. These points are typically the solutions to the equation obtained by setting the inequality's quadratic expression to zero.
In our example, the critical points \(x = -6\) and \(x = 4\) are determined from the factors \((x + 6)\) and \((x - 4)\). They divide the number line into distinct intervals where the inequality could switch from true to false or vice versa.
Testing regions around these critical points helps determine the sign of the inequality in each interval. The concept of critical points is crucial in solving inequalities as they guide us in testing and identifying valid solution intervals. Critical points make it easier to understand and visualize the behavior of the inequality across the number line.