Question

value of k if the equation x×x- 2(k+1)x + k×k where is equal to zero has equal roots

Answer 1

X²- 2(k+1)x + k² = 0

For equal roots,

B²-4ac = 0

(-2(k+1))² -4(k²) = 0

4(k+1)²-4k² = 0

(K+1)² - k² = 0

(K+1-k) (k+1+k)

K= -1/2

Answer 2

since x×x-2(k+1)+1=0 has real and equal roots. so discriminant must be 0.
so D=b^2-4ac=0
(2k+1)^2-4×1×k=0
4k^2+1+4k-4k=0
4k^2+1=0
k^2=-1/4

k=-1/2 answer
hope it helps