Question

X2-2(k+1)x+k2=0 find k for real equal roots

Answer 1

Answer:

to have real equal roots,

D=0

b²-4ac=0

here a=1 b=-2(k+1) c=k²

[-2(k+1)]²-4(1)(k²)=0

[4(k²+1+2k)]-4k²=0

4k²+4+8k-4k²=0

4+8k=0

8k=-4

k=-1/2

Answer 2

Answer:

-1/2 is the answer pl s mark me