Yes, this is an easy problem. I was able to solve it in my head. The graphs intersect in one point when k = 0 or 4 and in two points when k is negative of greater then 4. No intersecting happens when k is 1, 2 or 3. For k ≥ 0 we need to fulfil x/2 < √x in order to make the graph not intersect, because x + y = k is a line with slope of -1 and xy = k is hyperbole in first and third quadrant with the x and y axes being the asymptotes. So, we can compare the point on the line y = x at which these graphs intersect it and that's were we get the x/2 < √x from. Since √5 < 5/2 and the derivate of x/2 is constant 1/2 while the derivative of √x is 1/(2√5) < 1/2 at x=5 and goes to zero when x goes to infinity, if is clear x/2 > √x for x ≥ 5. We only need to check k = 0, 1, 2, 3 or 4. Clearly k = 0 or k = 4 don't work leaving us k = 1, 2 or 3. For k < 0, x + y = k is still a line with slope of -1, but xy = k becomes hyperbole in the second and fourth quadrants intersecting the line always once in both of these quadrants.
@@electricgamer_yt4753 k=2 is an integer, is it not? I was providing an example of a pair of complex numbers that when added or multiplied give the same integer result.
Poor doubt. Once think what is difference between argand plane and cartesian plane. One have it's axis as imaginary and real,while later have as x and y
