Answer :

let p(x) = x2 + kx + k

x =


=


=


k (k - 4) > 0 k (- ∞, 0) (4, ∞)


Here k lies in two intervals, therefore we need to consider both the intervals separately.


Case1:


When k (- ∞, 0)


i.e. k < 0


we know that in a quadratic equation p(x) = ax2 + bx + c, if either a > 0, c < 0 or a < 0, c > 0, then the zeroes of the polynomial are of opposite signs.


Here a = 1 > 0, b = k < 0 and c = k < 0


both zeroes are of opposite signs


Case2:


When k (4, ∞)


i.e. k > 0


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


Here a = 1 > 0, b = k > 0 and c = k > 0


both the zeroes are negative


Hence, by both cases, both the zeroes cannot be positive.

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