If `a+b+c=0`, then, the equation `3ax^(2)+2bx+c=0` has , in the interval (0,1).

Options

A. at least one root
B. at most one root
C. no root
D. none of these

Explanation

To solve the problem, we need to analyze the given quadratic equation \(3ax^2 + 2bx + c = 0\) under the condition that \(a + b + c = 0\). We will show that this equation has at least one root in the interval (0, 1).

### Step-by-Step Solution:

1. **Define the Function**:
Let \(f(x) = 3ax^2 + 2bx + c\). We need to find the roots of this function in the interval (0, 1).

2. **Evaluate the Function at the Endpoints**:
We will evaluate \(f(x)\) at \(x = 0\) and \(x = 1\).

- For \(x = 0\):
\[ f(0) = 3a(0)^2 + 2b(0) + c = c \]

- For \(x = 1\):
\[ f(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c \]

3. **Substitute the Condition \(a + b + c = 0\)**:
From the condition \(a + b + c = 0\), we can express \(c\) as:
\[ c = -a - b \]
Now substitute this into \(f(1)\):
\[ f(1) = 3a + 2b - a - b = 2a + b \]

4. **Evaluate \(f(0)\) and \(f(1)\)**:
- \(f(0) = c = -a - b\)
- \(f(1) = 2a + b\)

5. **Check the Signs of \(f(0)\) and \(f(1)\)**:
We need to analyze the signs of \(f(0)\) and \(f(1)\):
- If \(f(0) < 0\) and \(f(1) > 0\) or vice versa, then by the Intermediate Value Theorem, there is at least one root in the interval (0, 1).

6. **Consider Cases**:
- **Case 1**: If \(a > 0\) and \(b < 0\), then:
- \(f(0) = -a - b < 0\) (since \(-b > a\))
- \(f(1) = 2a + b > 0\) (since \(2a > -b\))
- **Case 2**: If \(a < 0\) and \(b > 0\), then:
- \(f(0) = -a - b > 0\)
- \(f(1) = 2a + b < 0\)

7. **Conclusion**:
In both cases, we find that \(f(0)\) and \(f(1)\) have opposite signs. Therefore, by the Intermediate Value Theorem, there is at least one root of the equation \(3ax^2 + 2bx + c = 0\) in the interval (0, 1).

### Final Answer:
The equation \(3ax^2 + 2bx + c = 0\) has at least one root in the interval (0, 1).

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