Problem 96 - College Algebra Essentials

Question

Solve each equation. $$ \left|x^{2}+6 x+1\right|=8 $$

Step-by-Step Solution

Answer

The solution to the equation \(\mid x^2+6x+1 \mid = 8\) is \(x=-7, -1\).

Step 1: Set Up Two Equations

Because the result of absolute value operation is always positive, the quadratic equation inside the absolute value tags can equal to 8 or -8. Thereby, the given equation splits into 2 equations: \(x^2+6x+1 = 8\) and \(x^2+6x+1 = -8\).

Step 2: Simplify Each Equation

Subtract 8 from both sides of the first equation to isolate the quadratic on the left side of the equation, and add 8 to both sides of the second equation to achieve the same. The results are \(x^2+6x-7 = 0\) and \(x^2+6x+9 = 0\).

Step 3: Solve First Quadratic Equation

The first equation \(x^2+6x-7 = 0\) can be solved by factoring, but it doesn't factor nicely, so the quadratic formula should be used. The quadratic formula is \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\), in this case, a=1, b=6, c=-7. So the solutions are \(x = [-6 ± sqrt((6)^2 - 4*1*(-7))] / (2*1) = -1, -7\).

Step 4: Solve Second Quadratic Equation

The second equation \(x^2+6x+9 = 0\) factors nicely into \((x+3)(x+3) = 0\), thus the solution for this equation is \(x = -3\). However, because the square root of any real number is either positive or zero, if the number inside the absolute value tags is negative, there won't be any real solution. Hence, in this case, \(x=-3\) is not a valid solution.

Key Concepts

Understanding Absolute Value Equations

Absolute value equations involve expressions where the absolute value of a number or expression is set equal to another number. The absolute value, symbolized by vertical bars, represents the distance of a number from zero on a number line. Therefore, it is always a non-negative value.

This means that when solving equations like \( |x^2+6x+1|=8 \), the expression inside the absolute value could either equal the positive value (8) or its negative (-8).
Split the original equation into two separate equations:

• \( x^2 + 6x + 1 = 8 \)
• \( x^2 + 6x + 1 = -8 \)

This splitting gives us two potential equations to solve. By understanding absolute value equations, we recognize these steps ensure all possible real solutions are considered.

Factoring Quadratic Equations

Factoring is a method used to solve quadratic equations by expressing the equation as a product of its factors. The goal is to set an equation in standard form \( ax^2 + bx + c = 0 \) and express it as \( (px + q)(rx + s) = 0 \).

When solving the equation \( x^2+6x+9 = 0 \) through factoring, observe that it can be rewritten as \((x+3)(x+3)=0\). Both factors are the same, resulting in a solution \(x = -3\).
This method is efficient for quadratics that "factor nicely," meaning when the roots are integers or can be easily identified. In our example, however, the solution \(x = -3\) is not valid due to the nature of the absolute value equation, where the negative result would not match the equation's original positive requirement. This illustrates the importance of checking the roots’ validity when they originate from absolute values.

Utilizing the Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of any quadratic equation, particularly when the equation cannot be easily factored. This formula is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\]In the equation \( x^2 + 6x - 7 = 0 \), it doesn’t factor simply, so the quadratic formula is used:

• a = 1
• b = 6
• c = -7

Substitute these values into the quadratic formula to get:
\[x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times (-7)}}{2 \times 1}\\]Calculate the discriminant \( b^2 - 4ac \) which ensures the solutions are real numbers. Here, it gives positive results, resulting in two real solutions: \(x = -1\) and \(x = -7\).

Thus, the quadratic formula not only simplifies our solving process but also emphasizes cases where factoring might not readily apply. It's a crucial method ensuring comprehensive solutions beyond simple factoring.