The equation`x^2=1-abs(x-2)` has n real solutions, then

Options

A. n=0
B. n=1
C. n=2
D. n=3

Answer

To solve the equation \( x^2 = 1 - |x - 2| \), we will analyze both sides of the equation step by step.

### Step 1: Understand the Absolute Value
The expression \( |x - 2| \) can be broken down into two cases based on the value of \( x \).

1. **Case 1:** When \( x \geq 2 \), then \( |x - 2| = x - 2 \).
2. **Case 2:** When \( x < 2 \), then \( |x - 2| = 2 - x \).

### Step 2: Solve for Each Case

#### Case 1: \( x \geq 2 \)
In this case, we can substitute \( |x - 2| \) with \( x - 2 \):
\[ x^2 = 1 - (x - 2) \]
Simplifying this gives:
\[ x^2 = 1 - x + 2 \]
\[ x^2 + x - 3 = 0 \]
Now we can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]
\[ = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2} \]
Calculating the roots:
\[ x = \frac{-1 + \sqrt{13}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{13}}{2} \]
We need to check which of these roots satisfy \( x \geq 2 \):
- \( \frac{-1 + \sqrt{13}}{2} \) is approximately \( 1.30 \) (not valid).
- \( \frac{-1 - \sqrt{13}}{2} \) is negative (not valid).

Thus, **no solutions** from Case 1.

#### Case 2: \( x < 2 \)
In this case, we substitute \( |x - 2| \) with \( 2 - x \):
\[ x^2 = 1 - (2 - x) \]
Simplifying this gives:
\[ x^2 = 1 - 2 + x \]
\[ x^2 - x + 1 = 0 \]
Now we can find the discriminant to check for real solutions:
\[ D = b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, there are **no real solutions** from Case 2.

### Conclusion
Since both cases yield no real solutions, the total number of real solutions \( n = 0 \).

### Final Answer
The value of \( n \) is \( 0 \).

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