NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 2: Polynomials - Exercise 2.1
Question 7

Question. 7

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

(A)

both positive

(B)

both negative

(C)

one positive and one negative

(D)

both equal

Step 1: Recall the relations between coefficients and zeroes.

For quadratic polynomial \(ax^2 + bx + c\):

Sum of zeroes = \(-\dfrac{b}{a}\),

Product of zeroes = \(\dfrac{c}{a}\).

Step 2: Apply to the given polynomial.

Here, \(a = 1\), \(b = 99\), \(c = 127\).

So,

Sum = \(-\dfrac{99}{1} = -99\).

Product = \(\dfrac{127}{1} = 127\).

Step 3: Interpret the signs.

Since product \(127 > 0\), both zeroes must have the same sign.

Since sum \(-99 < 0\), the common sign must be negative.

Conclusion: Both zeroes are negative.