If `x^(2)+bx+x=0` has no real roots and `a+b+c lt 0` then

Options

A. `4a+2b+x gt 0`
B. `a-b+c=0`
C. `c lt 0`
D. `a+b+c=0`

Answer

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

### Step-by-Step Solution:

1. **Identify the Condition for No Real Roots**:
A quadratic equation \( ax^2 + bx + c = 0 \) has no real roots if its discriminant \( D < 0 \). The discriminant \( D \) for the equation \( x^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Since \( a = 1 \) in our case, the discriminant simplifies to:
\[ D = b^2 - 4c \]
Therefore, for the equation to have no real roots, we must have:
\[ b^2 - 4c < 0 \]
This implies:
\[ b^2 < 4c \]

2. **Using the Condition \( a + b + c < 0 \)**:
Given \( a + b + c < 0 \) and substituting \( a = 1 \), we have:
\[ 1 + b + c < 0 \]
Rearranging this gives:
\[ b + c < -1 \]

3. **Combining the Conditions**:
We now have two inequalities:
- \( b^2 < 4c \)
- \( b + c < -1 \)

From the second inequality, we can express \( c \) in terms of \( b \):
\[ c < -1 - b \]

Substituting this into the first inequality:
\[ b^2 < 4(-1 - b) \]
Simplifying this gives:
\[ b^2 < -4 - 4b \]
Rearranging leads to:
\[ b^2 + 4b + 4 < 0 \]
This can be factored as:
\[ (b + 2)^2 < 0 \]

4. **Analyzing the Result**:
The expression \( (b + 2)^2 < 0 \) is never true for real numbers since a square of a real number is always non-negative. Therefore, the only way for the original conditions to hold is if the quadratic has no real roots and \( a + b + c < 0 \) is satisfied.

5. **Conclusion**:
The conditions imply that \( b + 2 = 0 \) or \( b = -2 \) is the only case where the quadratic can be at the boundary of having no real roots, and hence \( c \) must be less than \( -1 \).

### Final Answer:
Thus, the conditions lead us to conclude that \( c < -1 \) and \( b = -2 \).