Consider the equation #0 = x^2 + 4x + 4#. We can solve this by factoring as a perfect square trinomial, so #0 = (x+ 2)^2-> x = -2 and -2#. Hence, there will be two identical solutions.
The discriminant of the quadratic equation (#b^2 - 4ac#) can be used to determine the number and the type of solutions. Since a quadratic equations roots are in fact its x intercepts, and a perfect square trinomial will have #2# equal, or #1# distinct solution, the vertex lies on the x axis. We can set the discriminant to 0 and solve:
#k^2 - (4 xx 1 xx 36) = 0#
#k^2 - 144 = 0#
#(k + 12)(k - 12) = 0#
#k = +-12#
So, k can either be #12# or #-12#.
Hopefully this helps!
