y = x^2 + a
y = ax
Set the y's equal
x^2 + a = ax rearrange as
x^2 - ax + a = 0
For this to have real solutions.......the discriminant must be ≥ 0
So
a^2 - 4a ≥ 0
a ( a - 4) ≥ 0
Setting each factor to 0 and solving for a produces the following solutions a = 0 and a = 4
So we have the following possible intervals that will produce solutions
(-inf, 0 ] or ( 0 , 4 ) or [ 4, inf )
If a is in the first interval, then a (a - 4) ≥ 0 so this interval produces a solution
If a is in the second interval, then a(a - 4) < 0.....so no solutions are found here
If a is in the third interval, then a (a - 4) ≥ 0 .....so this interval produces a solution
So....the solution intervals are
(-inf, 0 ] U [ 4, inf )
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