Question

Find the value of k for which the equation (3k+1)x^2+(k+1)x+k=0 has real and equal roots

Answer 1

Answer:

Step-by-step explanation:

real and equal roots

so d=0

b²= 4ac

(k+1)²= 4*( 3k+1)*k

k²+1 +2k =  4*( 3k²+k)

11k²+2k-1 =0

k= (-2±√48)÷22