Find all values of k for which the equation 3x2−(k+2)x+k−1=0 has no solutions.

there are solutions for all values of k

Step-by-step explanation:

In order for there to be no real solutions, the value of the discriminant must be negative. For this equation, the discriminant is ...

(-(k+2))^2 -4(3)(k-1)

We want to find values of k for which this is negative:

(k^2 +4k +4) -12k +12 < 0

k^2 -8k +16 < 0

(k -4)^2 < 0

A square is never negative, so there are no values of k that result in the equation having no solutions.

answer: 590

step-by-step explanation:

the answer would be c:   ax + b = ax + c

step-by-step explanation: