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 given: it has no real roots and \( a + c > 0 \).

### Step-by-step Solution:

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

**Hint**: Remember that the discriminant determines the nature of the roots of a quadratic equation.

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

**Hint**: Consider how the signs of \( a \) and \( c \) affect the overall expression.

3. **Analyzing the Signs of \( a \) and \( c \)**:
Since \( a + c > 0 \), we can infer that both \( a \) and \( c \) must be either both positive or both negative. However, since \( a \) is the coefficient of \( x^2 \) in a quadratic equation, it cannot be negative (as it would open downwards and potentially allow for real roots). Thus, we conclude:
\[ a > 0 \quad \text{and} \quad c > 0 \]

**Hint**: Think about the implications of \( a \) being positive on the shape of the parabola.

4. **Relating \( a \), \( b \), and \( c \)**:
Given that \( b^2 < 4ac \) and both \( a \) and \( c \) are positive, we can say that \( 4ac \) is also positive. Therefore, \( b^2 \) must be less than a positive number, which implies:
\[ b^2 < 4ac \implies b^2 \text{ can take any real value, but it must be bounded.} \]

**Hint**: Consider the implications of \( b \) being either positive or negative.

5. **Conclusion**:
Since \( a > 0 \) and \( c > 0 \), and \( b^2 < 4ac \), we conclude that:
- The quadratic function \( ax^2 + bx + c \) is always positive for all \( x \) (since it opens upwards and does not cross the x-axis).
- Therefore, \( a - b + c > 0 \) must also hold true, as shown in the video transcript.

**Final Result**: The conditions lead us to conclude that \( a > 0 \), \( c > 0 \), and \( a - b + c > 0 \).