If `f(x)` be a quadratic polynomial such that `f(x)=0` has a root 3 and `f(2)+f(-1)=0` then other root lies in

Options

A. `(-1,0)`
B. `(0,1)`
C. `(-2,1)`
D. `(1,2)

Answer

To solve the problem, we need to find the interval in which the other root of the quadratic polynomial \( f(x) \) lies, given that one root is 3 and \( f(2) + f(-1) = 0 \).

### Step-by-Step Solution

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

2. **Use the Root Information**:
Since 3 is a root, we can substitute \( x = 3 \) into the polynomial:
\[ f(3) = a(3^2) + b(3) + c = 0 \implies 9a + 3b + c = 0 \quad \text{(Equation 1)} \]

3. **Use the Given Condition \( f(2) + f(-1) = 0 \)**:
First, calculate \( f(2) \):
\[ f(2) = a(2^2) + b(2) + c = 4a + 2b + c \]
Now calculate \( f(-1) \):
\[ f(-1) = a(-1^2) + b(-1) + c = a - b + c \]
Now, substitute these into the condition:
\[ f(2) + f(-1) = (4a + 2b + c) + (a - b + c) = 0 \]
Simplifying this gives:
\[ 5a + b + 2c = 0 \quad \text{(Equation 2)} \]

4. **Solve the System of Equations**:
We have two equations:
- \( 9a + 3b + c = 0 \) (Equation 1)
- \( 5a + b + 2c = 0 \) (Equation 2)

From Equation 1, we can express \( c \) in terms of \( a \) and \( b \):
\[ c = -9a - 3b \]
Substitute this expression for \( c \) into Equation 2:
\[ 5a + b + 2(-9a - 3b) = 0 \]
Simplifying gives:
\[ 5a + b - 18a - 6b = 0 \implies -13a - 5b = 0 \implies 13a = -5b \implies b = -\frac{13}{5}a \]

5. **Substitute Back to Find \( c \)**:
Substitute \( b \) back into the expression for \( c \):
\[ c = -9a - 3\left(-\frac{13}{5}a\right) = -9a + \frac{39}{5}a = -\frac{45}{5}a + \frac{39}{5}a = -\frac{6}{5}a \]

6. **Find the Other Root**:
The roots of the quadratic polynomial can be expressed as:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]
Here, \( \alpha = 3 \), \( b = -\frac{13}{5}a \), and \( c = -\frac{6}{5}a \):
\[ \alpha + \beta = -\left(-\frac{13}{5}\right) = \frac{13}{5} \implies 3 + \beta = \frac{13}{5} \implies \beta = \frac{13}{5} - 3 = \frac{13}{5} - \frac{15}{5} = -\frac{2}{5} \]

7. **Determine the Interval for \( \beta \)**:
The value of \( \beta = -\frac{2}{5} \) is approximately -0.4. This value lies between -1 and 0.

### Conclusion
Thus, the other root \( \beta \) lies in the interval \( (-1, 0) \).