The number of solutions of the equation
`sqrt(x^(2))-sqrt((x-1)^(2))+sqrt((x-2)^(2))=sqrt(5)` is

Options

A. `0`
B. `1`
C. `2`
D. More than `2`

Explanation

To solve the equation \( \sqrt{x^2} - \sqrt{(x-1)^2} + \sqrt{(x-2)^2} = \sqrt{5} \), we will express the square roots in terms of absolute values and analyze the equation in different cases based on the value of \( x \).

### Step 1: Rewrite the equation using absolute values

We know that \( \sqrt{a^2} = |a| \). Therefore, we can rewrite the equation as:

\[ |x| - |x-1| + |x-2| = \sqrt{5} \]

### Step 2: Analyze the cases based on the value of \( x \)

We will consider three cases based on the critical points where the expressions inside the absolute values change, which are \( x = 0 \), \( x = 1 \), and \( x = 2 \).

#### Case 1: \( x > 2 \)

In this case, all expressions are positive:

\[ |x| = x, \quad |x-1| = x-1, \quad |x-2| = x-2 \]

Substituting these into the equation gives:

\[ x - (x-1) + (x-2) = \sqrt{5} \]

Simplifying this:

\[ x - x + 1 + x - 2 = \sqrt{5} \]
\[ x - 1 = \sqrt{5} \]
\[ x = 1 + \sqrt{5} \]

Since \( 1 + \sqrt{5} \) is approximately \( 3.24 \), which is greater than \( 2 \), this solution is valid.

#### Case 2: \( x < 0 \)

In this case, all expressions are negative:

\[ |x| = -x, \quad |x-1| = -x + 1, \quad |x-2| = -x + 2 \]

Substituting these into the equation gives:

\[ -x - (-x + 1) + (-x + 2) = \sqrt{5} \]

Simplifying this:

\[ -x + x - 1 - x + 2 = \sqrt{5} \]
\[ -x + 1 = \sqrt{5} \]
\[ x = 1 - \sqrt{5} \]

Since \( 1 - \sqrt{5} \) is approximately \( -1.24 \), which is less than \( 0 \), this solution is valid.

#### Case 3: \( 0 \leq x < 1 \)

In this case:

\[ |x| = x, \quad |x-1| = -x + 1, \quad |x-2| = -x + 2 \]

Substituting these into the equation gives:

\[ x - (-x + 1) + (-x + 2) = \sqrt{5} \]

Simplifying this:

\[ x + x - 1 - x + 2 = \sqrt{5} \]
\[ x + 1 = \sqrt{5} \]
\[ x = \sqrt{5} - 1 \]

Since \( \sqrt{5} - 1 \) is approximately \( 1.24 \), which is not in the interval \( [0, 1) \), this solution is invalid.

#### Case 4: \( 1 \leq x < 2 \)

In this case:

\[ |x| = x, \quad |x-1| = x - 1, \quad |x-2| = -x + 2 \]

Substituting these into the equation gives:

\[ x - (x - 1) + (-x + 2) = \sqrt{5} \]

Simplifying this:

\[ x - x + 1 - x + 2 = \sqrt{5} \]
\[ 3 - x = \sqrt{5} \]
\[ x = 3 - \sqrt{5} \]

Since \( 3 - \sqrt{5} \) is approximately \( 0.76 \), which is in the interval \( [1, 2) \), this solution is invalid.

### Conclusion

The valid solutions we found are:

1. \( x = 1 + \sqrt{5} \) (from Case 1)
2. \( x = 1 - \sqrt{5} \) (from Case 2)

Thus, the number of solutions to the equation is **2**.