1.

(ix)(3k + 1)2 + 2 (k + 1) x + k = 0

Answer»

If it has equal roots the , D=0from this equation ,a=(3k+1) b= 2(k+1) c= k+1atq,D=b^2 -4ac0= 4(k^2 + 1 + 2k) - 4(3k+1)(k+1)0= -8k(k +1)k=0 k=-1



Discussion

No Comment Found