Answer :

let p(x) = x^{2} + 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) = ax^{2} + 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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