If `4a+2b+c=0` , then the equation `3ax^(2)+2bx+c=0` has at least one real root lying in the interval

Options

A. (0,1)
B. (1,2)
C. (0,2)
D. none of these

Answer

To solve the problem step by step, we need to analyze the given conditions and apply the Mean Value Theorem (MVT) to the polynomial function.

### Step-by-Step Solution:

1. **Identify the Given Equation**:
We are given the equation:
\[ 4a + 2b + c = 0 \]
We need to analyze the polynomial:
\[ 3ax^2 + 2bx + c = 0 \]

2. **Define the Function**:
Let:
\[ f(x) = 3ax^2 + 2bx + c \]
We want to find if there is at least one real root of this function in the interval [0, 2].

3. **Evaluate the Function at the Endpoints**:
Calculate \( f(0) \) and \( f(2) \):
- For \( x = 0 \):
\[ f(0) = 3a(0)^2 + 2b(0) + c = c \]
- For \( x = 2 \):
\[ f(2) = 3a(2)^2 + 2b(2) + c = 12a + 4b + c \]

4. **Substitute the Condition**:
From the given condition \( 4a + 2b + c = 0 \), we can express \( c \) in terms of \( a \) and \( b \):
\[ c = -4a - 2b \]
Substitute this into \( f(2) \):
\[ f(2) = 12a + 4b + (-4a - 2b) = 12a + 4b - 4a - 2b = 8a + 2b \]

5. **Evaluate \( f(0) \) and \( f(2) \)**:
Now we have:
\[ f(0) = c = -4a - 2b \]
\[ f(2) = 8a + 2b \]

6. **Check the Values**:
We need to check the signs of \( f(0) \) and \( f(2) \):
- If \( f(0) = -4a - 2b \) and \( f(2) = 8a + 2b \), we want to see if these two values have opposite signs:
- For \( f(0) \) and \( f(2) \) to have opposite signs, we need:
\[ (-4a - 2b)(8a + 2b) < 0 \]
- This means there is at least one root in the interval [0, 2] by the Intermediate Value Theorem.

7. **Conclusion**:
Since \( f(0) \) and \( f(2) \) have opposite signs, we can conclude that there is at least one real root of the equation \( 3ax^2 + 2bx + c = 0 \) in the interval [0, 2].

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

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