If `f(x)=ax^2+bx+c` and `g(x)=-ax^2+bx+c`, where `acne0`, the new equation f(x).g(x)=0 has n real roots. Find the minimum value of n.

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to analyze the equations given:

1. **Definitions**:
- \( f(x) = ax^2 + bx + c \)
- \( g(x) = -ax^2 + bx + c \)

2. **Setting Up the Equation**:
We want to find the roots of the equation \( f(x) \cdot g(x) = 0 \). This means we need to find the roots of both \( f(x) = 0 \) and \( g(x) = 0 \).

3. **Finding Roots of \( f(x) \)**:
The roots of \( f(x) = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The condition for \( f(x) \) to have real roots is:
\[ b^2 - 4ac \geq 0 \]

4. **Finding Roots of \( g(x) \)**:
Similarly, for \( g(x) = 0 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4(-a)c}}{2(-a)} = \frac{-b \pm \sqrt{b^2 + 4ac}}{-2a} \]
The condition for \( g(x) \) to have real roots is:
\[ b^2 + 4ac \geq 0 \]

5. **Analyzing the Conditions**:
Given that \( ac \neq 0 \), we can have two cases:
- **Case 1**: \( ac > 0 \)
- In this case, both \( b^2 - 4ac \) and \( b^2 + 4ac \) can be positive, leading to 2 real roots from both equations.
- **Case 2**: \( ac < 0 \)
- Here, \( b^2 - 4ac \) can still be positive, giving 2 real roots from \( f(x) \), while \( g(x) \) will also yield 2 real roots.

6. **Combining the Roots**:
From both cases, we can conclude:
- If \( f(x) \) has 2 real roots and \( g(x) \) has 2 real roots, the total roots from \( f(x) \cdot g(x) = 0 \) will be at least 2.
- If either \( f(x) \) or \( g(x) \) has complex roots, the other must have real roots, ensuring at least 2 real roots overall.

7. **Conclusion**:
Therefore, the minimum value of \( n \), the number of real roots of the equation \( f(x) \cdot g(x) = 0 \), is:
\[ \boxed{2} \]