The zeroes of the quadratic polynomial `x^(2)+99x +127` are

Options

A. both positive
B. both negative
C. one positive and one negative
D. both equal

Explanation

To find the zeroes of the quadratic polynomial \( x^2 + 99x + 127 \), we will follow these steps:

### Step 1: Write the polynomial and set it equal to zero
We start with the polynomial:
\[ x^2 + 99x + 127 = 0 \]

### Step 2: Identify coefficients
From the polynomial, we identify the coefficients:
- \( a = 1 \) (coefficient of \( x^2 \))
- \( b = 99 \) (coefficient of \( x \))
- \( c = 127 \) (constant term)

### Step 3: Calculate the discriminant
The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = 99^2 - 4 \cdot 1 \cdot 127 \]
Calculating \( 99^2 \):
\[ 99^2 = 9801 \]
Calculating \( 4 \cdot 1 \cdot 127 \):
\[ 4 \cdot 127 = 508 \]
Now substituting these values back into the discriminant formula:
\[ D = 9801 - 508 = 9293 \]

### Step 4: Calculate the square root of the discriminant
Next, we find the square root of the discriminant:
\[ \sqrt{D} = \sqrt{9293} \approx 96.4 \]

### Step 5: Use the quadratic formula to find the zeroes
The zeroes of the polynomial can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values of \( b \), \( \sqrt{D} \), and \( a \):
\[ x = \frac{-99 \pm 96.4}{2 \cdot 1} \]
This gives us two equations to solve:
1. \( x_1 = \frac{-99 + 96.4}{2} \)
2. \( x_2 = \frac{-99 - 96.4}{2} \)

### Step 6: Calculate the first zero
Calculating \( x_1 \):
\[ x_1 = \frac{-99 + 96.4}{2} = \frac{-2.6}{2} = -1.3 \]

### Step 7: Calculate the second zero
Calculating \( x_2 \):
\[ x_2 = \frac{-99 - 96.4}{2} = \frac{-195.4}{2} = -97.7 \]

### Conclusion
The zeroes of the quadratic polynomial \( x^2 + 99x + 127 \) are:
\[ x_1 = -1.3 \quad \text{and} \quad x_2 = -97.7 \]

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