If `c gt0` and `4a+clt2b` then `ax^(2)-bx+c=0` has a root in the interval

Options

A. (a) (0,2)
B. (b) (2,4)
C. (c) (0,1)
D. (d) (-2,0)

Answer

To solve the problem, we need to determine the interval in which the quadratic equation \( ax^2 - bx + c = 0 \) has a root, given the conditions \( c > 0 \) and \( 4a + c < 2b \).

### Step-by-Step Solution:

1. **Define the Function**:
We start by defining the function based on the quadratic equation:
\[ f(x) = ax^2 - bx + c \]

2. **Evaluate \( f(0) \)**:
Calculate \( f(0) \):
\[ f(0) = a(0)^2 - b(0) + c = c \]
Since \( c > 0 \), we have:
\[ f(0) > 0 \]

3. **Evaluate \( f(2) \)**:
Next, we calculate \( f(2) \):
\[ f(2) = a(2)^2 - b(2) + c = 4a - 2b + c \]
From the given condition \( 4a + c < 2b \), we can rearrange it to:
\[ 4a + c - 2b < 0 \implies f(2) < 0 \]

4. **Analyze the Signs of \( f(0) \) and \( f(2) \)**:
We have determined that:
- \( f(0) > 0 \) (positive)
- \( f(2) < 0 \) (negative)

5. **Apply the Intermediate Value Theorem**:
Since \( f(x) \) is a continuous function (as it is a polynomial), and it changes sign from positive at \( x = 0 \) to negative at \( x = 2 \), by the Intermediate Value Theorem, there must be at least one root in the interval \( (0, 2) \).

6. **Conclusion**:
Therefore, we conclude that the quadratic equation \( ax^2 - bx + c = 0 \) has at least one root in the interval:
\[ (0, 2) \]

### Final Answer:
The root lies in the interval \( (0, 2) \).