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

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -