Question

find the value of k,equation has real and equal roots(3k+1)X2+2(k+1)X+3=0​

Answer 1

Step-by-step explanation:

${b}^{2} - 4ac = 0$

$a = 3k + 1 \\ b = 2(k + 1)$

C=3

1. (2k+2)^2-4(3k+1)(3)=0

=k^2-7k-2=0,