Let `a gt 0, b gt 0 and c lt0.` Then, both the roots of the equation `ax^(2) +bx+c=0`

Options

A. are real and negative
B. have negative real parts
C. have positive real parts
D. None of these

Answer

To solve the problem, we need to analyze the quadratic equation given by:

\[ ax^2 + bx + c = 0 \]

where \( a > 0 \), \( b > 0 \), and \( c < 0 \). We will determine the nature of the roots based on these conditions.

### Step 1: Calculate the Discriminant

The discriminant \( D \) of the quadratic equation is given by:

\[ D = b^2 - 4ac \]

### Step 2: Analyze the Discriminant

Since \( a > 0 \) and \( c < 0 \), the term \( -4ac \) will be positive because:

- \( a \) is positive,
- \( c \) is negative,
- thus \( -4ac > 0 \).

Now, since \( b^2 \) is also positive (as \( b > 0 \)), we can conclude that:

\[ D = b^2 - 4ac > 0 \]

This indicates that the quadratic equation has two distinct real roots.

### Step 3: Calculate the Roots

The roots of the quadratic equation can be calculated using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} \]

Since we have established that \( D > 0 \), we can write:

\[ x_1 = \frac{-b + \sqrt{D}}{2a} \]
\[ x_2 = \frac{-b - \sqrt{D}}{2a} \]

### Step 4: Determine the Sign of the Roots

1. **For \( x_1 \)**:
- Since \( b > 0 \) and \( \sqrt{D} > 0 \), the term \( -b + \sqrt{D} \) could be positive or negative. However, since \( \sqrt{D} \) is less than \( b \) (as we will see in the next point), \( x_1 \) will be negative.

2. **For \( x_2 \)**:
- The term \( -b - \sqrt{D} \) is clearly negative since both \( -b \) and \( -\sqrt{D} \) are negative. Thus, \( x_2 \) will also be negative.

### Conclusion

Both roots \( x_1 \) and \( x_2 \) are negative. Therefore, the answer to the question is that both roots of the equation \( ax^2 + bx + c = 0 \) are negative.