What is the product of all the solutions of \((\mathrm{x} + 2)^2 = |\mathrm{x} + 2|\)?

1. Translate the problem requirements: We need to find all values of x that satisfy the equation \((\mathrm{x} + 2)^2 = |\mathrm{x} + 2|\), then multiply all those solution values together to get the final product.
2. Analyze when squared expressions equal absolute value expressions: Determine the conditions under which \(\mathrm{y}^2 = |\mathrm{y}|\) holds true by considering the behavior of both sides for positive and negative values.
3. Apply the condition to our specific expression: Use the general condition to find the specific values of \((\mathrm{x} + 2)\) that work, then solve for the corresponding x values.
4. Calculate the product of all solutions: Multiply all the x values we found to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have the equation \((\mathrm{x} + 2)^2 = |\mathrm{x} + 2|\), and we need to find all values of x that make this equation true. Once we have all those x values, we'll multiply them together to get our final answer.

Think of it this way: we're looking for numbers where if you square something, you get the same result as taking its absolute value. This might seem strange at first, but let's explore when this can happen.

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

2. Analyze when squared expressions equal absolute value expressions

Let's think about this in everyday terms. When does \(\mathrm{y}^2\) equal \(|\mathrm{y}|\)?

Imagine you have some number y. Let's try a few examples:

- If \(\mathrm{y} = 3\), then \(\mathrm{y}^2 = 9\) and \(|\mathrm{y}| = 3\). Since \(9 \neq 3\), this doesn't work.
- If \(\mathrm{y} = 1\), then \(\mathrm{y}^2 = 1\) and \(|\mathrm{y}| = 1\). Since \(1 = 1\), this works!
- If \(\mathrm{y} = 0\), then \(\mathrm{y}^2 = 0\) and \(|\mathrm{y}| = 0\). Since \(0 = 0\), this works too!
- If \(\mathrm{y} = -2\), then \(\mathrm{y}^2 = 4\) and \(|\mathrm{y}| = 2\). Since \(4 \neq 2\), this doesn't work.

From these examples, we can see a pattern emerging. Let's think about this more systematically:

For any number y, we need \(\mathrm{y}^2 = |\mathrm{y}|\). We can rearrange this as \(\mathrm{y}^2 - |\mathrm{y}| = 0\), or \(|\mathrm{y}|(|\mathrm{y}| - 1) = 0\).

In plain English: either the absolute value is zero, or the absolute value minus 1 is zero.

This means either \(|\mathrm{y}| = 0\) or \(|\mathrm{y}| = 1\).

So \(\mathrm{y} = 0\), \(\mathrm{y} = 1\), or \(\mathrm{y} = -1\).

Process Skill: CONSIDER ALL CASES - Systematically examining when the equation holds

3. Apply the condition to our specific expression

Now we know that \(\mathrm{y}^2 = |\mathrm{y}|\) only when y equals 0, 1, or -1.

In our problem, \(\mathrm{y} = (\mathrm{x} + 2)\). So we need:

- \(\mathrm{x} + 2 = 0\), which gives us \(\mathrm{x} = -2\)
- \(\mathrm{x} + 2 = 1\), which gives us \(\mathrm{x} = -1\)
- \(\mathrm{x} + 2 = -1\), which gives us \(\mathrm{x} = -3\)

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

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

Perfect! All three solutions work.

4. Calculate the product of all solutions

Now we multiply all our x values together:

Product = \((-2) \times (-1) \times (-3)\)

Let's do this step by step:
First: \((-2) \times (-1) = 2\) (negative times negative gives positive)
Then: \(2 \times (-3) = -6\) (positive times negative gives negative)

Therefore, the product of all solutions is -6.

Final Answer

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

We can double-check this makes sense: we found three solutions \((\mathrm{x} = -2, -1, -3)\), and their product \((-2)(-1)(-3) = -6\), which matches our answer choices perfectly.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the fundamental condition

Students often struggle to recognize that the equation \((\mathrm{x} + 2)^2 = |\mathrm{x} + 2|\) requires them to find when a squared expression equals an absolute value expression. They may try to solve this by expanding the left side or treating the absolute value as a simple algebraic expression, rather than understanding that this only occurs for specific values (0, 1, -1) of the expression inside.

2. Incomplete case analysis

Some students may recognize they need to consider cases for the absolute value but fail to systematically determine all possible values. They might only consider when \(\mathrm{x} + 2 \geq 0\) and \(\mathrm{x} + 2 < 0\) without fully analyzing when \(\mathrm{y}^2 = |\mathrm{y}|\) is satisfied, missing the critical insight that this only happens when \(|\mathrm{y}| = 0\) or \(|\mathrm{y}| = 1\).

Errors while executing the approach

1. Sign errors in arithmetic

When solving the individual equations \((\mathrm{x} + 2 = 0, \mathrm{x} + 2 = 1, \mathrm{x} + 2 = -1)\), students may make basic arithmetic errors, particularly with negative numbers. For example, they might incorrectly solve \(\mathrm{x} + 2 = -1\) as \(\mathrm{x} = 1\) instead of \(\mathrm{x} = -3\).

2. Incomplete verification

Students may find the correct x-values but fail to verify them properly in the original equation. They might make computational errors when checking solutions like substituting \(\mathrm{x} = -3\) and incorrectly calculating \((\mathrm{x} + 2)^2\) or \(|\mathrm{x} + 2|\), leading them to discard valid solutions.

Errors while selecting the answer

1. Multiplication sequence errors

When calculating the product \((-2) \times (-1) \times (-3)\), students often make sign errors. They might correctly calculate \((-2) \times (-1) = 2\) but then incorrectly compute \(2 \times (-3)\) as +6 instead of -6, forgetting that positive times negative gives negative.

2. Misreading the final requirement

Some students may correctly find all three solutions \((\mathrm{x} = -2, -1, -3)\) but misunderstand what the question is asking for. They might add the solutions instead of multiplying them, getting -6 by coincidence, or they might forget to include all three solutions in their final calculation.