Solving |x^2 + 8x + 12| = 5: A Step-by-Step Guide

Hey guys! Today, we're diving into a fun little math problem involving absolute values and quadratic equations. Specifically, we're going to solve the equation |x² + 8x + 12| = 5. Don't worry; it's not as scary as it looks! We'll break it down step by step, so you can easily follow along and master this type of problem. Let's get started!

Understanding Absolute Value Equations

Before we jump into the specifics of our equation, let's quickly recap what absolute value means. The absolute value of a number is its distance from zero, regardless of direction. For example, |3| = 3 and |-3| = 3. This means that when we have an absolute value equation like |x| = a, we need to consider two possibilities: x = a and x = -a. This is because both 'a' and '-a' are 'a' units away from zero. So, when solving an absolute value equation, we always split it into two separate equations.

Now, relating this understanding to our problem |x² + 8x + 12| = 5, it implies that the expression inside the absolute value, i.e., x² + 8x + 12, can be either 5 or -5. This is because both 5 and -5 have an absolute value of 5. Hence, we will create and solve two quadratic equations derived from this understanding. We'll solve each quadratic equation separately, and the solutions to both equations will be the solutions to our original absolute value equation. Remember, each quadratic equation can have up to two real solutions, so we might end up with as many as four solutions for 'x'. Keep this in mind as we proceed to the next section where we break down the problem into two distinct quadratic equations.

Breaking Down the Equation

As we discussed, the absolute value equation |x² + 8x + 12| = 5 leads to two separate equations:

1. x² + 8x + 12 = 5
2. x² + 8x + 12 = -5

Let's tackle each of these equations one at a time.

Solving x² + 8x + 12 = 5

First, we need to rearrange the equation to get it into the standard quadratic form, which is ax² + bx + c = 0. To do this, we subtract 5 from both sides of the equation:

x² + 8x + 12 - 5 = 0 x² + 8x + 7 = 0

Now we have a standard quadratic equation. There are several ways to solve this, including factoring, completing the square, or using the quadratic formula. In this case, factoring is the easiest approach. We're looking for two numbers that multiply to 7 and add up to 8. Those numbers are 7 and 1. So, we can factor the quadratic as follows:

(x + 7)(x + 1) = 0

To find the solutions for x, we set each factor equal to zero:

x + 7 = 0 or x + 1 = 0

Solving these simple equations gives us:

x = -7 or x = -1

So, the solutions for the first equation are x = -7 and x = -1. Make sure to hold onto these solutions as we move on to the next equation. Remember, these are only part of the full solution set for the original absolute value equation.

Solving x² + 8x + 12 = -5

Now, let's move on to the second equation: x² + 8x + 12 = -5. Again, we need to rearrange this equation into the standard quadratic form by adding 5 to both sides:

x² + 8x + 12 + 5 = 0 x² + 8x + 17 = 0

This quadratic equation doesn't factor as easily as the first one. So, we'll use the quadratic formula to find the solutions. The quadratic formula is:

x = (-b ± √(b² - 4ac)) / (2a)

In our equation, a = 1, b = 8, and c = 17. Plugging these values into the quadratic formula, we get:

x = (-8 ± √(8² - 4 * 1 * 17)) / (2 * 1) x = (-8 ± √(64 - 68)) / 2 x = (-8 ± √(-4)) / 2

Notice that we have a negative number under the square root. This means that the solutions will be complex numbers. Specifically, √(-4) = 2i, where 'i' is the imaginary unit (i² = -1). So, we can rewrite the solutions as:

x = (-8 ± 2i) / 2 x = -4 ± i

Therefore, the solutions for the second equation are x = -4 + i and x = -4 - i. These are complex solutions, which means they are not real numbers.

Combining the Solutions

Alright, we've solved both equations! Let's gather all our solutions. From the first equation, x² + 8x + 12 = 5, we found the real solutions x = -7 and x = -1. From the second equation, x² + 8x + 12 = -5, we found the complex solutions x = -4 + i and x = -4 - i.

So, the complete set of solutions for the original absolute value equation |x² + 8x + 12| = 5 is:

x = -7, x = -1, x = -4 + i, and x = -4 - i

It's important to note that depending on the context of the problem, you might only be interested in the real solutions. In that case, you would only include x = -7 and x = -1 in your final answer. However, if the problem asks for all solutions, you need to include both the real and complex solutions.

Verification of the Solutions

To ensure accuracy, it's always a good idea to verify our solutions. Let's start by verifying the real solutions, x = -7 and x = -1, using the original equation |x² + 8x + 12| = 5.

Verifying x = -7

Substitute x = -7 into the original equation:

|(-7)² + 8(-7) + 12| = |49 - 56 + 12| = |5| = 5

The left side of the equation equals the right side, so x = -7 is indeed a solution.

Verifying x = -1

Substitute x = -1 into the original equation:

|(-1)² + 8(-1) + 12| = |1 - 8 + 12| = |5| = 5

Again, the left side of the equation equals the right side, confirming that x = -1 is also a solution.

Verifying Complex Solutions

Verifying the complex solutions requires a bit more care. Let's verify x = -4 + i:

|(-4 + i)² + 8(-4 + i) + 12| = |(16 - 8i - 1) - 32 + 8i + 12| = |-5| = 5

And now let's verify x = -4 - i:

|(-4 - i)² + 8(-4 - i) + 12| = |(16 + 8i - 1) - 32 - 8i + 12| = |-5| = 5

Both complex solutions also satisfy the original equation. This completes our verification process, affirming that all our solutions are correct.

Conclusion

So, there you have it! We've successfully solved the absolute value equation |x² + 8x + 12| = 5. We broke it down into two separate quadratic equations, solved each one, and combined the solutions. Remember the key steps:

1. Split the absolute value equation into two equations.
2. Rearrange each equation into standard quadratic form.
3. Solve each quadratic equation using factoring or the quadratic formula.
4. Combine all the solutions.
5. Verify all the solutions in the original equation.

With practice, you'll become a pro at solving these types of equations. Keep up the great work, and happy math-solving!