Question

For what value of k does the equation kt^2 +5t = 0 have exactly one real solution for t?

Answer 1

Hint:

ax^2+bx + c = 0

has exactly one solution for x when b^2 - 4ac = 0

Answer 2

kt^2 +5t = t(kt +5) = 0 already has t = 0 as a solution, regardless of k

Answer 3

...but since c=0 and b=5 no can do.

k=0 works