Question

The zeros of the quadratic polynomial \(x^2 + 99x + 127\) are :

• A. Both Positive
• B. Both Negative
• C. One Positive and one Negative
• D. Both equal

Hint:

For \(ax^2+bx+c\), analyze the sum (\(S = -b/a\)) and product (\(P = c/a\)) of zeros: Polynomial: \(x^2 + 99x + 127\) Here, \(a=1, b=99, c=127\). 1. Sum \(S = -99/1 = -99\) (Negative) 2. Product \(P = 127/1 = 127\) (Positive) Analysis:
Product \(P>0\): Means zeros have the same sign (both +ve or both -ve).
Sum \(S<0\):
If both were +ve, sum would be +ve. (Not our case)
If both are -ve, sum is -ve. (Matches our sum) Therefore, both zeros are negative. (Discriminant \(D = 99^2 - 4 \cdot 1 \cdot 127 = 9801 - 508 = 9293>0\), so real distinct roots exist).

Answer 1

The Correct Option is B

Concept: For a quadratic polynomial \(ax^2 + bx + c\), let \(\alpha\) and \(\beta\) be its zeros.
Sum of zeros: \(\alpha + \beta = -\frac{b}{a}\)
Product of zeros: \(\alpha\beta = \frac{c}{a}\)
Discriminant: \(D = b^2 - 4ac\). If \(D>0\), zeros are real and distinct. If \(D = 0\), zeros are real and equal. If \(D<0\), zeros are complex (not real). The signs of the sum and product of zeros can tell us about the signs of the zeros themselves. Step 1: Identify coefficients a, b, and c The given polynomial is \(x^2 + 99x + 127\). Comparing with \(ax^2 + bx + c\):
\(a = 1\)
\(b = 99\)
\(c = 127\) Step 2: Calculate the sum of the zeros (\(\alpha + \beta\)) \[ \alpha + \beta = -\frac{b}{a} = -\frac{99}{1} = -99 \] Since the sum of the zeros is negative (\(-99<0\)). Step 3: Calculate the product of the zeros (\(\alpha\beta\)) \[ \alpha\beta = \frac{c}{a} = \frac{127}{1} = 127 \] Since the product of the zeros is positive (\(127>0\)). Step 4: Analyze the signs of the zeros based on their sum and product
If the product \(\alpha\beta\) is positive, it means either both zeros are positive OR both zeros are negative.
If the product \(\alpha\beta\) is negative, it means one zero is positive and the other is negative. In our case, \(\alpha\beta = 127\) (positive). So, either both zeros are positive or both are negative. Now consider the sum \(\alpha + \beta = -99\) (negative).
If both zeros were positive, their sum would be positive. This contradicts \(\alpha + \beta = -99\).
If both zeros are negative, their sum would be negative. This is consistent with \(\alpha + \beta = -99\). Therefore, both zeros must be negative. Step 5: Check the discriminant (optional, to ensure real zeros) \(D = b^2 - 4ac = (99)^2 - 4(1)(127)\) \(99^2 = (100-1)^2 = 10000 - 200 + 1 = 9801\) \(4(1)(127) = 4 \times 127 = 508\) \(D = 9801 - 508 = 9293\). Since \(D = 9293>0\), the zeros are real and distinct. Since \(D \neq 0\), the zeros are not equal. Conclusion: The zeros are both negative. This matches option (2).