the number of non-zero solutions of the equation `x^2-5x-(sgn x)6=0` is.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( x^2 - 5x - \text{sgn}(x) \cdot 6 = 0 \) and find the number of non-zero solutions, we will analyze the equation based on the definition of the signum function, \( \text{sgn}(x) \).

### Step 1: Understand the Signum Function
The signum function, \( \text{sgn}(x) \), takes three values:
- \( \text{sgn}(x) = 1 \) when \( x > 0 \)
- \( \text{sgn}(x) = 0 \) when \( x = 0 \)
- \( \text{sgn}(x) = -1 \) when \( x < 0 \)

### Step 2: Case 1 - \( x > 0 \)
When \( x > 0 \), we have:
\[ \text{sgn}(x) = 1 \]
Substituting this into the equation:
\[ x^2 - 5x - 6 = 0 \]
This can be factored as:
\[ (x - 6)(x + 1) = 0 \]
Setting each factor to zero gives:
\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
Since we are only considering \( x > 0 \), the valid solution here is \( x = 6 \).

### Step 3: Case 2 - \( x = 0 \)
When \( x = 0 \):
\[ \text{sgn}(x) = 0 \]
Substituting this into the equation:
\[ x^2 - 5x = 0 \]
Factoring gives:
\[ x(x - 5) = 0 \]
Setting each factor to zero gives:
\[ x = 0 \quad \text{or} \quad x = 5 \]
However, since we are looking for non-zero solutions, neither of these values are valid.

### Step 4: Case 3 - \( x < 0 \)
When \( x < 0 \), we have:
\[ \text{sgn}(x) = -1 \]
Substituting this into the equation:
\[ x^2 - 5x + 6 = 0 \]
This can be factored as:
\[ (x - 3)(x - 2) = 0 \]
Setting each factor to zero gives:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
Since we are only considering \( x < 0 \), neither \( x = 3 \) nor \( x = 2 \) are valid solutions.

### Conclusion
After analyzing all cases, we find that the only non-zero solution is \( x = 6 \) from the first case. Therefore, the number of non-zero solutions of the equation is:

\[ \boxed{1} \]