If `ax^(2)+bx+c = 0` has no real roots and a, b, c `in` R such that `a + c gt 0`, then

Options

A. `a - b + c lt 0`
B. `a - b + c gt 0`
C. `a + c = b`
D. all of these

Explanation

To solve the problem, we need to analyze the quadratic equation \( ax^2 + bx + c = 0 \) under the conditions that it has no real roots and that \( a + c > 0 \).

### Step-by-Step Solution:

1. **Understanding the Condition for No Real Roots**:
A quadratic equation has no real roots if its discriminant is less than zero. The discriminant \( D \) for the equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
For the equation to have no real roots, we require:
\[ D < 0 \implies b^2 - 4ac < 0 \implies b^2 < 4ac \]

2. **Analyzing the Condition \( a + c > 0 \)**:
We are given that \( a + c > 0 \). This means that the sum of the coefficients \( a \) and \( c \) is positive.

3. **Relating \( a \), \( b \), and \( c \)**:
Since \( b^2 < 4ac \) and \( a + c > 0 \), we can infer that both \( a \) and \( c \) must have the same sign (either both positive or both negative) for their sum to be positive. However, since we are looking for a relation involving \( a \), \( b \), and \( c \), we will consider the case where both \( a \) and \( c \) are positive.

4. **Considering the Function \( f(x) = ax^2 + bx + c \)**:
Since \( a > 0 \) (as \( a + c > 0 \) and both \( a \) and \( c \) are positive), the graph of the quadratic opens upwards. Given that there are no real roots, the entire graph lies above the x-axis for all \( x \).

5. **Evaluating \( f(-1) \)**:
We can evaluate the function at \( x = -1 \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
Since the quadratic function \( f(x) \) is always greater than zero for all \( x \), we have:
\[ f(-1) > 0 \implies a - b + c > 0 \]

6. **Conclusion**:
Therefore, we conclude that:
\[ a - b + c > 0 \]

### Final Answer:
The relation between \( a \), \( b \), and \( c \) is:
\[ a - b + c > 0 \]