A bag contains 5 red counters, 4 blue counters and k k\,k green counters, where k k\,k is a positive integer. A counter is selected at random, replaced, and then a second counter is selected.
The probability that both counters are green is 14\displaystyle \frac{1}{4}41.
Show that k2−6k−27=0k^2 - 6k - 27 = 0k2−6k−27=0.
Hence find the value of kkk.