Determine one of the zeros of the quadratic realtion: y=(x-(k/2))^2-[(k-2)^2/4]

Question

I tried to do this by expanding and got to y=x^2-mx-k+1

How are you suppose to do this

Answer 1

you have

y = (x-a)^2 - b
so,
(x-a)^2 = b
x = a±√b
= (k/2)±√((k-2)^2/4)
= (k/2)±(k-2)/2
= k/2 + k/2 - 1 = k-1
or k/2 - k/2 + 1 = 1

check x=1:
(1-(k/2))^2 - (k-2)^2/4
(2-k)^2/4 - (k-2)^2/4
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