Problem 45 - College Algebra

Question

For the following exercises, solve for the unknown variable. $$ \left|x^{2}+2 x-36\right|=12 $$

Step-by-Step Solution

Answer

The solutions are \( x = -8, 6, -6, \) and \( 4 \).

Step 1: Setup Equation

The equation is given as \( |x^2 + 2x - 36| = 12 \). This means either \( x^2 + 2x - 36 = 12 \) or \( x^2 + 2x - 36 = -12 \). These correspond to two separate equations that we need to solve.

Step 2: Solve First Case

First, solve the equation \( x^2 + 2x - 36 = 12 \). Start by simplifying it to \( x^2 + 2x - 48 = 0 \). Now factorize the quadratic equation: \( (x + 8)(x - 6) = 0 \). Therefore, the solutions for this equation are \( x = -8 \) and \( x = 6 \).

Step 3: Solve Second Case

Next, solve the equation \( x^2 + 2x - 36 = -12 \). Simplify it to \( x^2 + 2x - 24 = 0 \). Factorize this quadratic equation to: \( (x + 6)(x - 4) = 0 \). Consequently, the solutions for this equation are \( x = -6 \) and \( x = 4 \).

Step 4: Combine Solutions

Combine all solutions obtained from the two cases to find \( x \). The possible values of \( x \) are \( x = -8, 6, -6, \) and \( 4 \).

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that involve terms up to the second degree, meaning they include an \( x^2 \) term. They are usually expressed in the standard form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Understanding quadratic equations is crucial since they're foundational to high school algebra and appear often in various math problems.
To solve quadratic equations, you can use several methods such as:

• Factoring
• The quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

In solving the original problem, recognizing that the expressions inside the absolute value symbols are quadratics helped us to set the equations for factoring. This step is essential for solving equations involving absolute values.

Factoring

Factoring is the process of breaking down an expression into simpler multiplicative factors that, when multiplied together, give the original expression. For quadratic equations, factoring involves finding two binomials that multiply to form the given quadratic.
For example, the quadratic equation from the problem can be factored like so:

• Take \( x^2 + 2x - 48 = 0 \) and rewrite it as \( (x + 8)(x - 6) = 0 \)
• This tells us the roots of the equation are \( x = -8 \) and \( x = 6 \)

Factoring is often the quickest way to solve simple quadratic equations, especially when the roots are integers. However, not all quadratics factor neatly, and other methods might be needed. But in this exercise, factoring allowed us to systematically find the possible values for \( x \).

Absolute Value Properties

Absolute value refers to the distance of a number from zero on the number line, regardless of direction, thus always being non-negative. In mathematical terms, the absolute value of a number \( x \) is denoted as \( |x| \).
When introduced in equations, absolute values can create two possible scenarios for the equation since both the positive and negative versions of a number have the same absolute value. Consider the equation: \( |x^2 + 2x - 36| = 12 \). Here, this means:

• \( x^2 + 2x - 36 = 12 \)
• \( x^2 + 2x - 36 = -12 \)

This characteristic of absolute values splits the problem into two separate quadratic equations, doubling the scenarios we must solve. This understanding is critical when working with absolute value equations.

Solution Sets

Solution sets refer to the set of all possible values that satisfy the original equation. After breaking down the given absolute value equation and solving each part, we collect all possible solutions for the variable.
In the exercise, after solving both quadratic scenarios derived from \( |x^2 + 2x - 36| = 12 \):

• The first solution set from \( x^2 + 2x - 48 = 0 \) is \( x = -8 \), \( x = 6 \)
• The second solution set from \( x^2 + 2x - 24 = 0 \) is \( x = -6 \), \( x = 4 \)

By combining these solutions, the complete solution set is \( \{-8, 6, -6, 4\} \). Understanding solution sets in algebra helps us ensure all possible scenarios have been considered and no potential answers have been left out.