Answer:
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.
