If a + b + c = 0 then the quadratic equation `3ax^(2) + 2bx + c = 0` has

Answer

To solve the problem, we need to analyze the quadratic equation \(3ax^2 + 2bx + c = 0\) given that \(a + b + c = 0\).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We have the condition \(a + b + c = 0\). This can be rearranged to express \(c\) in terms of \(a\) and \(b\):
\[ c = - (a + b) \]

2. **Substituting \(c\) into the Quadratic Equation**:
Substitute \(c\) into the quadratic equation:
\[ 3ax^2 + 2bx + (- (a + b)) = 0 \]
This simplifies to:
\[ 3ax^2 + 2bx - a - b = 0 \]

3. **Using Rolle's Theorem**:
We define a function \(\phi(x) = ax^3 + bx^2 + cx\). By differentiating, we find:
\[ \phi'(x) = 3ax^2 + 2bx + c \]
We need to evaluate \(\phi(0)\) and \(\phi(1)\):
- \(\phi(0) = 0\)
- \(\phi(1) = a + b + c = 0\) (given)

4. **Applying Rolle's Theorem**:
Since \(\phi(0) = 0\) and \(\phi(1) = 0\), by Rolle's theorem, there exists at least one \( \lambda \) in the interval \( (0, 1) \) such that:
\[ \phi'(\lambda) = 0 \]
This implies:
\[ 3a\lambda^2 + 2b\lambda + c = 0 \]

5. **Conclusion**:
Since we have found a value \(\lambda\) in the interval \( (0, 1) \) such that the derivative \(\phi'(\lambda) = 0\), it indicates that the quadratic equation \(3ax^2 + 2bx + c = 0\) has at least one real root in the interval \( (0, 1) \). Therefore, the correct option is:
- **Real and different roots of which at least one lies in \(0\) and \(1\)**.