Question

Find value of k for which the given equation has real and equal roots x^2-2x (1+3k)+7 (3+2k)=0

Answer 1

x²-2x(1+3k)+7(3+2k)=0

x²-x(2+6k)+7(3+2k)=0

here a = 1, b= -(2+6k) and c = 7(3+2k)

Now, D= b²-4ac

(6k+2)²-4(1)x7(3+2k)

36k²+4+24k-84-56k

36k²-32k-80

Now, for equal roots D = 0

36k²-32k-80 = 0

9K²-8K-20=0

9K²-18K+10K-20=0

9K(K-2)+10(K-2)=0

(9K+10)(K-2)=0

K= -10/9 or k = 2