The zeros of the quadratic polynomial $x^2 +99x+127$ are:

Question

Accepted Answer

To determine the nature of the zeros of the quadratic polynomial $x^2 + 99x + 127$, we can use the properties of quadratic equations and the quadratic formula. The quadratic formula is given by:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For the polynomial $x^2 + 99x + 127$, the coefficients are $a = 1$, $b = 99$, and $c = 127$.

1. **Calculate the Discriminant:**

The discriminant ($D$) of a quadratic equation $ax^2 + bx + c$ is given by:

$$
   D = b^2 - 4ac
   $$

Substituting the values, we get:

$$
   D = 99^2 - 4 \times 1 \times 127 = 9801 - 508 = 9293
   $$

Since the discriminant is positive ($D > 0$), the quadratic equation has two distinct real roots.

2. **Determine the Sign of the Roots:**

The sum of the roots of the quadratic equation is given by $-\frac{b}{a}$, and the product of the roots is given by $\frac{c}{a}$.

- **Sum of the roots:** $-\frac{b}{a} = -\frac{99}{1} = -99$
   - **Product of the roots:** $\frac{c}{a} = \frac{127}{1} = 127$

Since the sum of the roots is negative ($-99$) and the product of the roots is positive ($127$), both roots must be negative. This is because the only way for two numbers to have a positive product and a negative sum is if both numbers are negative.

3. **Explanation of Options:**

- **Option 1 (Both positive):** This is incorrect because if both roots were positive, their sum would be positive, not negative.
   
   - **Option 2 (Both negative):** This is correct because the negative sum and positive product indicate that both roots are negative.
   
   - **Option 3 (Both equal):** This is incorrect because the discriminant is positive, indicating two distinct roots, not equal roots.
   
   - **Option 4 (One positive and one negative):** This is incorrect because if one root were positive and the other negative, the product of the roots would be negative, not positive.

Therefore, the correct answer is Option 2: Both roots are negative.