The zeroes of the quadratic polynomial x^{2} + 99x + 127 are

let p(x) = x^{2} + 99x + 127

x

=

=

=

= - 2.6/2, - 195.4/2

⇒ x = - 1.3, - 97.7

Both the zeroes are negative

OR

We know, in quadratic polynomial if the coefficients of the terms are of the same sign, then the zeroes of the polynomial are negative.

i.e. if either a > 0, b > 0 and c > 0 or a < 0, b < 0 and c < 0, then both zeroes are negative

So here, a = 1 > 0, b = 99 > 0 and c = 127 > 0

⇒ The zeroes are negative.

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