The zeroes of the quadratic polynomial x2 + 99x + 127 are
let p(x) = x2 + 99x + 127
x![]()
= ![]()
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= - 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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