`{x in R: |x-2|= x^(2)}=`

Options

A. {-1, 2}
B. {1, 2}
C. {-1, -2}
D. {1, -2}

Answer

To solve the equation \( |x - 2| = x^2 \), we will analyze it by considering two cases based on the definition of the absolute value.

### Step 1: Set up the cases based on the absolute value
The absolute value function \( |x - 2| \) can be expressed in two cases:
1. Case 1: \( x - 2 \) when \( x \geq 2 \)
2. Case 2: \( -(x - 2) \) when \( x < 2 \)

### Step 2: Solve Case 1 (\( x \geq 2 \))
For \( x \geq 2 \):
\[ |x - 2| = x - 2 \]
Thus, the equation becomes:
\[ x - 2 = x^2 \]
Rearranging gives:
\[ x^2 - x + 2 = 0 \]

### Step 3: Calculate the discriminant for Case 1
To determine the nature of the roots, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 \]
Since \( D < 0 \), there are no real solutions in this case.

### Step 4: Solve Case 2 (\( x < 2 \))
For \( x < 2 \):
\[ |x - 2| = -(x - 2) = -x + 2 \]
Thus, the equation becomes:
\[ -x + 2 = x^2 \]
Rearranging gives:
\[ x^2 + x - 2 = 0 \]

### Step 5: Calculate the discriminant for Case 2
Now, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (1)^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9 \]
Since \( D > 0 \), there are two distinct real solutions.

### Step 6: Factor the quadratic equation for Case 2
We can factor the quadratic equation:
\[ x^2 + 2x - x - 2 = 0 \]
This can be factored as:
\[ (x + 2)(x - 1) = 0 \]
Thus, the solutions are:
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 7: Check the validity of the solutions
- For \( x = -2 \): This satisfies \( x < 2 \).
- For \( x = 1 \): This also satisfies \( x < 2 \).

### Final Solution
The solution set is:
\[ \{ -2, 1 \} \]