3p^2+2p= – (k+4)

Question

Find all values of k for which the equation has one real solution.3p2+2p=–(k+4)Write your answer starting with k,followed by an equals sign or inequality symbol (for example, k<5).Reduce all fractions.

Answer 1

3p^2 + 2p = w

p = -2/6 +/- sqr(4 -4(3)(-w))

when b^2 -4ac = 0, there's only one solution, b=2, a=3, c=w = k+4

4+12w = 0

w = -4/12 =-1/3 = k+4

k+4 = -1/3

k = -1/3 -4 = -1/3 -12/3

k = -13/3 = - 4 1/3