If `P(x)=ax^2+bx+c`, `Q(x)=-ax^2+dx+c` where `ac!=0` then `P(x).Q(x)=0` has

Options

A. four real roots
B. two real roots
C. four imaginary roots
D. none of these

Answer

To solve the problem, we need to analyze the product of the two polynomials \( P(x) \) and \( Q(x) \) and determine the conditions under which the equation \( P(x) \cdot Q(x) = 0 \) has real roots.

### Step-by-Step Solution:

1. **Define the Polynomials**:
We have:
\[ P(x) = ax^2 + bx + c \]
\[ Q(x) = -ax^2 + dx + c \]

2. **Condition for Real Roots**:
For \( P(x) \) to have real roots, the discriminant must be non-negative:
\[ D_P = b^2 - 4ac \geq 0 \]
For \( Q(x) \) to have real roots, we also need its discriminant to be non-negative:
\[ D_Q = d^2 - 4(-a)c = d^2 + 4ac \geq 0 \]

3. **Analyze the Conditions**:
We know that \( ac \neq 0 \). We will analyze the two cases based on the sign of \( ac \).

- **Case 1**: If \( ac < 0 \):
- Since \( ac \) is negative, \( 4ac \) will also be negative. Thus:
\[ D_P = b^2 - 4ac > 0 \quad \text{(since subtracting a negative is like adding)} \]
\[ D_Q = d^2 + 4ac > 0 \quad \text{(since \( 4ac \) is negative, but \( d^2 \) is non-negative)} \]
Therefore, both \( P(x) \) and \( Q(x) \) have real roots.

- **Case 2**: If \( ac > 0 \):
- Here, \( 4ac \) is positive. Thus:
\[ D_P = b^2 - 4ac \geq 0 \]
\[ D_Q = d^2 + 4ac > 0 \quad \text{(since both terms are positive)} \]
Again, both \( P(x) \) and \( Q(x) \) have real roots.

4. **Conclusion**:
In both cases, whether \( ac < 0 \) or \( ac > 0 \), the product \( P(x) \cdot Q(x) = 0 \) will have at least two real roots, one from each polynomial.

Thus, we conclude that \( P(x) \cdot Q(x) = 0 \) has at least two real roots.