Let `f(x)=x^(2)+ax+b`, where `a, b in R`. If `f(x)=0` has all its roots imaginary, then the roots of `f(x)+f'(x)+f''(x)=0` are

Options

A. real and distinct
B. imaginary
C. equal
D. rational and equal

Answer

To solve the problem, we need to analyze the quadratic function \( f(x) = x^2 + ax + b \) and determine the conditions under which its roots are imaginary. Then, we will find the roots of the equation \( f(x) + f'(x) + f''(x) = 0 \).

### Step-by-Step Solution:

1. **Identify the Condition for Imaginary Roots:**
The roots of the quadratic equation \( f(x) = 0 \) are imaginary if the discriminant is less than zero. The discriminant \( D \) for the quadratic \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
For our function \( f(x) = x^2 + ax + b \), we have:
\[ D = a^2 - 4(1)(b) = a^2 - 4b \]
To have imaginary roots, we require:
\[ a^2 - 4b < 0 \quad \text{(1)} \]

2. **Calculate the Derivatives:**
Next, we need to find the first and second derivatives of \( f(x) \):
\[ f'(x) = 2x + a \]
\[ f''(x) = 2 \]

3. **Formulate the New Equation:**
We need to solve the equation:
\[ f(x) + f'(x) + f''(x) = 0 \]
Substituting \( f(x) \), \( f'(x) \), and \( f''(x) \):
\[ (x^2 + ax + b) + (2x + a) + 2 = 0 \]
Simplifying this, we combine like terms:
\[ x^2 + (a + 2)x + (b + a + 2) = 0 \]

4. **Determine the Discriminant of the New Quadratic:**
Now, we need to find the discriminant of this new quadratic equation:
\[ D' = (a + 2)^2 - 4(1)(b + a + 2) \]
Expanding this:
\[ D' = (a + 2)^2 - 4(b + a + 2) \]
\[ = a^2 + 4a + 4 - 4b - 4a - 8 \]
\[ = a^2 - 4b - 4 \]

5. **Analyze the Discriminant:**
From (1), we know that \( a^2 - 4b < 0 \). Therefore:
\[ a^2 - 4b - 4 < -4 \quad \text{(since } a^2 - 4b < 0\text{)} \]
This implies:
\[ D' < 0 \]
Thus, the quadratic \( x^2 + (a + 2)x + (b + a + 2) = 0 \) also has imaginary roots.

### Conclusion:
The roots of the equation \( f(x) + f'(x) + f''(x) = 0 \) are imaginary.