Let f(x) be a quadratic polynomial such that `f(-2) + f(3)= 0`. If one of the roots of f(x) = 0 is –1, then the sum of the roots of f(x) = 0 is equal to:

Options

A. `(11)/(3)`
B. `(7)/(3)`
C. `(13)/(3)`
D. `(14)/(3)`

Answer

To solve the problem, we need to find the sum of the roots of the quadratic polynomial \( f(x) \) given that \( f(-2) + f(3) = 0 \) and one of the roots is \( -1 \).

### Step-by-Step Solution:

1. **Assume the Form of the Quadratic Polynomial:**
Let \( f(x) = ax^2 + bx + c \).

2. **Calculate \( f(-2) \) and \( f(3) \):**
- For \( f(-2) \):
\[ f(-2) = a(-2)^2 + b(-2) + c = 4a - 2b + c \]
- For \( f(3) \):
\[ f(3) = a(3)^2 + b(3) + c = 9a + 3b + c \]

3. **Set Up the Equation from the Given Condition:**
According to the problem, we have:
\[ f(-2) + f(3) = 0 \]
Substituting the expressions we found:
\[ (4a - 2b + c) + (9a + 3b + c) = 0 \]
Simplifying this gives:
\[ 13a + b + 2c = 0 \quad \text{(Equation 1)} \]

4. **Use the Root Information:**
Since one of the roots is \( -1 \), we know that:
\[ f(-1) = 0 \]
Thus:
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c = 0 \quad \text{(Equation 2)} \]

5. **Solve the System of Equations:**
We have two equations:
- \( 13a + b + 2c = 0 \) (Equation 1)
- \( a - b + c = 0 \) (Equation 2)

From Equation 2, we can express \( c \) in terms of \( a \) and \( b \):
\[ c = b - a \]

Substitute \( c \) into Equation 1:
\[ 13a + b + 2(b - a) = 0 \]
Simplifying this gives:
\[ 13a + b + 2b - 2a = 0 \implies 11a + 3b = 0 \]
Therefore:
\[ 3b = -11a \implies b = -\frac{11}{3}a \]

6. **Find the Sum of the Roots:**
The sum of the roots of a quadratic polynomial \( ax^2 + bx + c \) is given by:
\[ -\frac{b}{a} \]
Substituting \( b = -\frac{11}{3}a \):
\[ \text{Sum of the roots} = -\frac{-\frac{11}{3}a}{a} = \frac{11}{3} \]

### Final Answer:
The sum of the roots of \( f(x) = 0 \) is \( \frac{11}{3} \).