How you you work out this question?

k<0 or k>4

Step-by-step explanation:

In the generic equation ax² + bx + c = 0, there are two solutions only if discriminant D>0 where D = b²-4ac.

Mapping your equation on the generic means that:

a = k

b = k

c = 1

So when we have to solve D>0 we solve:

k² - 4k > 0

k(k-4) > 0

roots are k=0 and k=4

k² - 4k > 0 when k<0 or k>4

k < 0 or k > 4

Step-by-step explanation:

kx² + kx + 1 = 0

Two solutions:

b²-4ac > 0

(k)²-4(k)(1) > 0

k² - 4k > 0

k(k-4) > 0

k < 0 or k > 4

your answer is 125y5