Problem 27 - College Algebra

Question

Solve each quadratic inequality in Exercises \(1-28\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \left|x^{2}+2 x-36\right|>12 $$

Step-by-Step Solution

Answer

The solution set of the inequality in interval notation is: \((-∞, -8) ∪ (-4, 6) ∪ (6, ∞)\)

Step 1: Remove the Absolute Value

An absolute value of a number is always positive, so |x^2 + 2x - 36 | can either be \(x^2 + 2x - 36\) or \(-(x^2 + 2x - 36)\). Thus, the given absolute value inequality is equivalent to two separate inequalities: \(x^2 + 2x - 36 > 12\) and \(-(x^2 + 2x - 36) > 12\).

Step 2: Simplify the Inequalities

Subtract 12 from both sides of both inequalities to simplify them: \(x^2 + 2x - 48 > 0\) and \(-(x^2 + 2x - 24) > 0\). This results into two further inequalities: \(x^2 + 2x - 48 > 0\) and \(-x^2 - 2x + 24 < 0\).

Step 3: Factor the Quadratics

By factoring, the inequalities become: \((x+8)(x-6) > 0\) and \((x+4)(x-6) < 0\). You have two critical points for each inequality.

Step 4: Solve the Inequalities

Determine the roots of the inequalities are by setting each factor equal to zero and solving for x. For the first one the roots will be at -8 and 6. Throw these values on a number line, break the number line into intervals, plug in representative from those intervals and you find that the solution for this inequality is \((-∞, -8) ∪ (6, ∞)\). For the second inequality, roots will be at -4 and 6. Using the similar method, you find that its solution is \((-4, 6)\).

Step 5: Combine the Solution Sets

Since the absolute value inequality is a compound 'or' inequality, solutions for either of the two inequalities are valid. Therefore, combine both previous solutions to get the final solution in interval notation: \((-∞, -8) ∪ (-4, 6) ∪ (6, ∞)\).

Key Concepts

Understanding Absolute Value

The concept of absolute value is crucial when dealing with inequalities like \(\left|x^{2}+2x-36\right|>12\). Absolute value represents the distance from zero on a number line, making it always non-negative. In other words, \(|x|\) looks for the size of \(x\) disregarding its sign.

To solve equations involving absolute values, this involves evaluating two separate scenarios:

• \(x^{2}+2x-36 > 12\)
• \(x^{2}+2x-36 < -12\)

This approach captures both distance directions on the number line. Solving separately helps in understanding all possible solutions.

Factoring Quadratics

Factoring is an efficient method to solve quadratic expressions like \(x^2 + 2x - 48\). Factoring breaks down a quadratic into simpler binomial expressions. Consider the expression \(x^2 + 2x - 48\). To factor it, identify two numbers that multiply to \(-48\) and add up to \(2\):

• These numbers are \(8\) and \(-6\), giving us \((x+8)(x-6)\).

The quadratic \(x^2 + 2x - 24\) could be rewritten as \((x+4)(x-6)\). The factorization process reveals the critical points where the expression can change signs on the number line.

Using Interval Notation

Interval notation is a way of representing sets of numbers, especially beneficial when addressing solutions to inequalities. It describes the range of values making the inequality true.

For example, you determine solutions like \((-\infty, -8)\) or \((6, \infty)\). This shows solutions that are either less than \(-8\) or greater than \(6\). The use of brackets is essential (round brackets for non-inclusive ranges and square brackets if the endpoint is included). Each solution segment from inequalities can be expressed in interval notation for clarity.

Visualizing on the Real Number Line

The real number line is a visual tool for better understanding inequalities. After determining the critical points such as \(-8, -4,\) and \(6\), position these on the number line.

By doing this, one can quickly identify intervals where each inequality holds true. Testing points in the intervals will show which satisfies the inequalities. For instance, the interval \((-\infty, -8)\) implies values less than \(-8\), while \((6, \infty)\) covers values greater than \(6\). Graphing these intervals visually depicts the solution set, making the solution comprehensive and clear.