The product of all the solutions of the equation`(x-2)^(2)-3|x-2|+2=0` is

Options

A. 2
B. -4
C. 0
D. none of these

Answer

To solve the equation \((x-2)^{2} - 3|x-2| + 2 = 0\), we will consider two cases based on the definition of the absolute value.

### Step 1: Define Cases for Absolute Value
The expression \(|x-2|\) can be defined in two cases:
1. Case 1: \(x - 2 \geq 0\) (i.e., \(x \geq 2\))
2. Case 2: \(x - 2 < 0\) (i.e., \(x < 2\))

### Step 2: Solve Case 1 (\(x \geq 2\))
In this case, \(|x-2| = x-2\). Substituting this into the equation gives:
\[ (x-2)^{2} - 3(x-2) + 2 = 0 \]
Let \(t = x - 2\). Then the equation becomes:
\[ t^{2} - 3t + 2 = 0 \]
Now, we can factor this quadratic equation:
\[ (t - 1)(t - 2) = 0 \]
Thus, we find:
\[ t = 1 \quad \text{or} \quad t = 2 \]
Substituting back for \(x\):
- If \(t = 1\), then \(x - 2 = 1 \Rightarrow x = 3\)
- If \(t = 2\), then \(x - 2 = 2 \Rightarrow x = 4\)

### Step 3: Solve Case 2 (\(x < 2\))
In this case, \(|x-2| = -(x-2) = 2 - x\). Substituting this into the equation gives:
\[ (x-2)^{2} - 3(2-x) + 2 = 0 \]
This simplifies to:
\[ (x-2)^{2} + 3x - 6 + 2 = 0 \Rightarrow (x-2)^{2} + 3x - 4 = 0 \]
Again, let \(t = x - 2\):
\[ t^{2} + 3(t + 2) - 4 = 0 \Rightarrow t^{2} + 3t + 6 - 4 = 0 \Rightarrow t^{2} + 3t + 2 = 0 \]
Factoring gives:
\[ (t + 1)(t + 2) = 0 \]
Thus, we find:
\[ t = -1 \quad \text{or} \quad t = -2 \]
Substituting back for \(x\):
- If \(t = -1\), then \(x - 2 = -1 \Rightarrow x = 1\)
- If \(t = -2\), then \(x - 2 = -2 \Rightarrow x = 0\)

### Step 4: Collect All Solutions
From both cases, we have the solutions:
- From Case 1: \(x = 3, 4\)
- From Case 2: \(x = 1, 0\)

### Step 5: Calculate the Product of All Solutions
The solutions are \(0, 1, 3, 4\). The product of these solutions is:
\[ 0 \times 1 \times 3 \times 4 = 0 \]

### Final Answer
The product of all the solutions of the equation is \(\boxed{0}\).

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