i was told in this way
derivative of curve = derivate of tangent line at intersection
y' = 2x (derivative of curve)
y = 2 (derivative of tangent line)
2x = 2
x = 1
when x = 1 ,, y = 2(x) --> y = 2(1) --> y = 2
tangent point (1,2)
y = x^2 + k
2 = 1^2 + k
2 = 1 + k
k = 1
that was the way i was told 
lol i wanted to know what my sir did actually ...
Your teacher used the properties of gradients. You are told in the question that the line is a
tangent to the curve. At one point on a curve, there can only be ONE tangent. By logic, the tangent on the curve at the point of contact with the line y = 2x, imperatively has to be the line y = 2x itself.
Your teacher found the gradient of the line y = 2x, and the gradient of the curve (that is, the tangent) at the point he assumes to be (x, y), where the line and the curve touch. Since the line y = 2x is itself the tangent, both gradients must be the same. That's what your teacher compared to get the value of x, then y. And finally, he replaces to get k.
That's a simpler method.
Discriminant, as Astar said, is b^2 - 4ac. Refer to your Quadratic Formula for finding roots, the value of b^2 - 4ac, the discriminant, determines whether you will have real and distinct roots (b^2 - 4ac > 0), equal ones (b^2 - 4ac = 0), or imaginary ones (b^2 - 4ac < 0).
A tangent touches a curve at ONLY ONE point. Naturally, you should get only one value of x on solving both equations. Again referring to the Quadratic Formula, x will have only one value if
b^2 - 4ac = 0, i.e.
x= -b/2a.
That's the method Mr Paul used.
