find the least value of k for which the equation x^2-2kx+k^2=3+x has real roots

Answer 1

the discriminant must be non-negative. So,

x^2-2kx+k^2 = 3+x
x^2 - (2k+1)x + (k^2-3) = 0

the discriminant is

(2k+1)^2 - 4(1)(k^2-3)
= 4k^2+4k+1 - 4k^2+12
= 4k-11

so, 4k-11 >= 0
k >= 11/4

for k= 11/4, we have

x^2 - 11/2 x + 121/16 = 3 + x
x^2 - 13/2 x + 169/16 = 0
(x - 13/4)^2 = 0

So, at k = 11/4, we have two equal roots. Any smaller value of k will cause the discriminant to be negative, producing complex roots. You can prove that for yourself by letting

k = 11/4 - h

for some positive value of h.

Answer 2

i believe it should13/4
as 4k+1+12>=0
4k+13>=0
k>=13/4
correct me if I'm wrong