Question

For what value of k the given quadratic equation has real and equal roots:
x^2-2(k+1)x+k^2​

Answer 1

Answer:

$\huge\underline\mathcal\red{Answer}$

x²-2(k+1)x+k²

b²-4ac

(2(k+1)²-4(1)(k²)

4(k²+2k+1)-4k²

4k²+8k+4-4k²

8k+4=0.........D=0 in real and equal

k=-4/8

k=-1/2 OK friend

Answer 2

b²-4ac

(2(k+1)²-4(1)(k²)

4(k²+2k+1)-4k²

4k²+8k+4-4k²

8k+4=0.........D=0 in real and equal

k=-4/8

k=-1/2