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

Options

A. `a ( a + b + c) gt 0`
B. `c -c ( a - b - c) gt 0`
C. `c + c ( a- b - c) gt 0`
D. `c (a - b - c) gt 0`

Answer

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

### 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 \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
For the equation to have no real roots, we require:
\[ D < 0 \implies b^2 - 4ac < 0 \]

2. **Analyzing the Sign of \( a \)**:
The coefficient \( a \) can be either positive or negative. We will analyze both cases.

- **Case 1: \( a > 0 \)**:
If \( a \) is positive, the parabola opens upwards. For the parabola to not intersect the x-axis (i.e., have no real roots), it must lie entirely above the x-axis. This implies:
\[ f(x) = ax^2 + bx + c > 0 \quad \text{for all } x \]
In particular, evaluating at \( x = 1 \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c > 0 \]

- **Case 2: \( a < 0 \)**:
If \( a \) is negative, the parabola opens downwards. For it to not intersect the x-axis, it must lie entirely below the x-axis. Thus:
\[ f(x) = ax^2 + bx + c < 0 \quad \text{for all } x \]
Evaluating at \( x = 1 \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c < 0 \]

3. **Combining the Results**:
In both cases, we can conclude that:
- If \( a > 0 \), then \( a + b + c > 0 \).
- If \( a < 0 \), then \( a + b + c < 0 \).

However, since we are looking for a condition that holds true irrespective of the sign of \( a \), we can state:
\[ a(a + b + c) > 0 \]
This implies that the product of \( a \) and \( (a + b + c) \) is positive.

### Conclusion:
Thus, the condition that must hold true if the quadratic equation \( ax^2 + bx + c = 0 \) has no real roots is:
\[ a(a + b + c) > 0 \]