If `|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|` then the set of values of x is

Options

A. `(-oo , 0] cup [ 3,00)`
B. `[0,1/2] cup [2,3]`
C. `(-oo,0]cup [1/2,2] cup [3,oo)`
D. `[0,2] cup [3,oo)`

Explanation

To solve the equation \( |x^2 - 2x + 2| - |2x^2 - 5x + 2| = |x^2 - 3x| \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation:
\[ |x^2 - 2x + 2| = |2x^2 - 5x + 2| + |x^2 - 3x| \]

### Step 2: Analyze the expressions inside the absolute values
Next, we need to analyze the expressions inside the absolute values. We will find the roots of each quadratic expression to determine the intervals where they change sign.

1. **For \(x^2 - 2x + 2\)**:
- The discriminant \(D = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4\) (no real roots).
- This expression is always positive since it opens upwards.

2. **For \(2x^2 - 5x + 2\)**:
- The discriminant \(D = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9\).
- Roots: \(x = \frac{5 \pm 3}{4} \Rightarrow x = 2 \text{ and } x = \frac{1}{2}\).

3. **For \(x^2 - 3x\)**:
- The roots are \(x(x - 3) = 0\) giving \(x = 0 \text{ and } x = 3\).

### Step 3: Determine intervals
The critical points are \(0, \frac{1}{2}, 2, 3\). We will analyze the sign of the expressions in the intervals:
- \( (-\infty, 0) \)
- \( (0, \frac{1}{2}) \)
- \( (\frac{1}{2}, 2) \)
- \( (2, 3) \)
- \( (3, \infty) \)

### Step 4: Test each interval
1. **Interval \( (-\infty, 0) \)**:
- \( |x^2 - 2x + 2| = x^2 - 2x + 2 \)
- \( |2x^2 - 5x + 2| = -(2x^2 - 5x + 2) \)
- \( |x^2 - 3x| = -(x^2 - 3x) \)

2. **Interval \( (0, \frac{1}{2}) \)**:
- \( |x^2 - 2x + 2| = x^2 - 2x + 2 \)
- \( |2x^2 - 5x + 2| = -(2x^2 - 5x + 2) \)
- \( |x^2 - 3x| = -(x^2 - 3x) \)

3. **Interval \( (\frac{1}{2}, 2) \)**:
- \( |x^2 - 2x + 2| = x^2 - 2x + 2 \)
- \( |2x^2 - 5x + 2| = 2x^2 - 5x + 2 \)
- \( |x^2 - 3x| = -(x^2 - 3x) \)

4. **Interval \( (2, 3) \)**:
- \( |x^2 - 2x + 2| = x^2 - 2x + 2 \)
- \( |2x^2 - 5x + 2| = 2x^2 - 5x + 2 \)
- \( |x^2 - 3x| = x^2 - 3x \)

5. **Interval \( (3, \infty) \)**:
- \( |x^2 - 2x + 2| = x^2 - 2x + 2 \)
- \( |2x^2 - 5x + 2| = 2x^2 - 5x + 2 \)
- \( |x^2 - 3x| = x^2 - 3x \)

### Step 5: Solve inequalities
For each interval, we will set up inequalities based on the signs of the expressions and solve them.

### Step 6: Combine solutions
After solving all the inequalities, we will combine the solutions to find the set of values of \(x\).

### Final Result
The solution set is:
\[ x \in [0, \frac{1}{2}] \cup [2, 3] \]