What is the product of all the solutions of x^2 + 4x + 7 = |x + 2| + 3?

1. Translate the problem requirements: We need to find all values of x that satisfy \(\mathrm{x^2 + 4x + 7 = |x + 2| + 3}\), then multiply those solutions together to get the final answer.
2. Simplify the equation structure: Rearrange to isolate terms and recognize that we can work with \(\mathrm{x^2 + 4x + 4 = |x + 2|}\), which factors as \(\mathrm{(x + 2)^2 = |x + 2|}\).
3. Apply absolute value case analysis: Since we have \(\mathrm{|x + 2|}\), consider two cases - when \(\mathrm{(x + 2) \geq 0}\) and when \(\mathrm{(x + 2) < 0}\), solving each case separately.
4. Solve each case systematically: For each case, substitute the appropriate expression for \(\mathrm{|x + 2|}\) and solve the resulting equation to find all valid solutions.
5. Calculate the product of solutions: Multiply all valid solutions together to arrive at the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked to do. We have an equation with an absolute value: \(\mathrm{x^2 + 4x + 7 = |x + 2| + 3}\). We need to find ALL the values of x that make this equation true, then multiply those solutions together.

Think of it like this: we're looking for x-values where the left side equals the right side. Once we find all such x-values, we multiply them to get our final answer.

Process Skill: TRANSLATE - Converting the problem statement into clear mathematical objectives

2. Simplify the equation structure

Before dealing with the absolute value, let's rearrange our equation to make it easier to work with.

Starting with: \(\mathrm{x^2 + 4x + 7 = |x + 2| + 3}\)

Subtract 3 from both sides: \(\mathrm{x^2 + 4x + 4 = |x + 2|}\)

Now here's something interesting! The left side \(\mathrm{x^2 + 4x + 4}\) is actually a perfect square. Let's see:
\(\mathrm{x^2 + 4x + 4 = (x + 2)^2}\)

So our equation becomes: \(\mathrm{(x + 2)^2 = |x + 2|}\)

This is much cleaner! We now have the square of something equaling the absolute value of that same something.

Process Skill: SIMPLIFY - Recognizing the perfect square pattern to make the problem more manageable

3. Apply absolute value case analysis

When we have an absolute value \(\mathrm{|x + 2|}\), we need to think about what this means. The absolute value gives us the distance from zero, so:

• If \(\mathrm{(x + 2)}\) is positive or zero, then \(\mathrm{|x + 2| = x + 2}\)
• If \(\mathrm{(x + 2)}\) is negative, then \(\mathrm{|x + 2| = -(x + 2)}\)

This means we have two cases to consider:

• Case 1: When \(\mathrm{x + 2 \geq 0}\) (which means \(\mathrm{x \geq -2}\))
• Case 2: When \(\mathrm{x + 2 < 0}\) (which means \(\mathrm{x < -2}\))

Process Skill: CONSIDER ALL CASES - Systematic analysis of absolute value conditions

4. Solve each case systematically

Case 1: When \(\mathrm{x \geq -2}\)
Here, \(\mathrm{|x + 2| = x + 2}\), so our equation becomes:
\(\mathrm{(x + 2)^2 = x + 2}\)

Rearranging: \(\mathrm{(x + 2)^2 - (x + 2) = 0}\)
Factoring: \(\mathrm{(x + 2)[(x + 2) - 1] = 0}\)
Simplifying: \(\mathrm{(x + 2)(x + 1) = 0}\)

This gives us: \(\mathrm{x + 2 = 0}\) or \(\mathrm{x + 1 = 0}\)
So: \(\mathrm{x = -2}\) or \(\mathrm{x = -1}\)

Now we check if these satisfy our case condition \(\mathrm{x \geq -2}\):

• \(\mathrm{x = -2}\): Yes, \(\mathrm{-2 \geq -2}\) ✓
• \(\mathrm{x = -1}\): Yes, \(\mathrm{-1 \geq -2}\) ✓

Both solutions are valid for Case 1.

Case 2: When \(\mathrm{x < -2}\)
Here, \(\mathrm{|x + 2| = -(x + 2)}\), so our equation becomes:
\(\mathrm{(x + 2)^2 = -(x + 2)}\)

Rearranging: \(\mathrm{(x + 2)^2 + (x + 2) = 0}\)
Factoring: \(\mathrm{(x + 2)[(x + 2) + 1] = 0}\)
Simplifying: \(\mathrm{(x + 2)(x + 3) = 0}\)

This gives us: \(\mathrm{x + 2 = 0}\) or \(\mathrm{x + 3 = 0}\)
So: \(\mathrm{x = -2}\) or \(\mathrm{x = -3}\)

Now we check if these satisfy our case condition \(\mathrm{x < -2}\):

• \(\mathrm{x = -2}\): No, -2 is not < -2 ✗
• \(\mathrm{x = -3}\): Yes, \(\mathrm{-3 < -2}\) ✓

Only \(\mathrm{x = -3}\) is valid for Case 2.

5. Calculate the product of solutions

From our case analysis, we found three valid solutions:

• From Case 1: \(\mathrm{x = -2}\) and \(\mathrm{x = -1}\)
• From Case 2: \(\mathrm{x = -3}\)

Let's verify these solutions work in the original equation:

For \(\mathrm{x = -2}\): \(\mathrm{(-2)^2 + 4(-2) + 7 = 4 - 8 + 7 = 3}\), and \(\mathrm{|-2 + 2| + 3 = 0 + 3 = 3}\) ✓
For \(\mathrm{x = -1}\): \(\mathrm{(-1)^2 + 4(-1) + 7 = 1 - 4 + 7 = 4}\), and \(\mathrm{|-1 + 2| + 3 = 1 + 3 = 4}\) ✓
For \(\mathrm{x = -3}\): \(\mathrm{(-3)^2 + 4(-3) + 7 = 9 - 12 + 7 = 4}\), and \(\mathrm{|-3 + 2| + 3 = 1 + 3 = 4}\) ✓

All solutions check out!

The product of all solutions = \(\mathrm{(-2) \times (-1) \times (-3) = -6}\)

Final Answer

The product of all solutions is -6, which corresponds to answer choice A.

Process Skill: APPLY CONSTRAINTS - Verifying that each solution satisfies both the equation and the case conditions

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "product of all solutions" means
Students often confuse this with finding just one solution or the sum of solutions. They may solve correctly but then add the solutions \(\mathrm{(-2) + (-1) + (-3) = -6}\) instead of multiplying them, accidentally getting the right answer through wrong reasoning.

2. Failing to recognize the need for absolute value case analysis
Many students see \(\mathrm{|x + 2|}\) and either ignore the absolute value entirely or try to solve it without considering that absolute values create two different cases based on whether the expression inside is positive or negative.

3. Not recognizing the perfect square pattern
Students may miss that \(\mathrm{x^2 + 4x + 4 = (x + 2)^2}\), leading them to work with a more complicated equation structure instead of the simplified form \(\mathrm{(x + 2)^2 = |x + 2|}\), making the problem much harder than necessary.

Errors while executing the approach

1. Incorrectly applying case conditions or forgetting to check validity
Students often solve both cases correctly but forget to verify that their solutions actually satisfy the case conditions (\(\mathrm{x \geq -2}\) for Case 1, \(\mathrm{x < -2}\) for Case 2). This leads to including invalid solutions like \(\mathrm{x = -2}\) from Case 2.

2. Sign errors when handling the negative absolute value case
In Case 2 where \(\mathrm{|x + 2| = -(x + 2)}\), students frequently make errors with the negative sign, either forgetting it entirely or applying it incorrectly when setting up the equation \(\mathrm{(x + 2)^2 = -(x + 2)}\).

3. Arithmetic mistakes in factoring or solving quadratic-like equations
Students may incorrectly factor expressions like \(\mathrm{(x + 2)^2 - (x + 2) = 0}\) or \(\mathrm{(x + 2)^2 + (x + 2) = 0}\), leading to wrong solutions or missing solutions entirely.

Errors while selecting the answer

1. Including invalid solutions in the final product
Even if students identify all potential solutions, they may forget to exclude those that don't satisfy the case conditions, leading them to calculate \(\mathrm{(-2) \times (-1) \times (-2) \times (-3) = 12}\) instead of the correct \(\mathrm{(-2) \times (-1) \times (-3) = -6}\).

2. Sign errors in the final multiplication
Students may correctly identify the three solutions (-2, -1, -3) but make a sign error when multiplying, getting +6 instead of -6, especially if they lose track of having an odd number of negative factors.