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This can be readily solved by first setting the parabolas equal to each other, or:
k x 2 − 4 x + 4 = 2 x 2 + 2 k x ;
or ( k + 2 ) x 2 − ( 4 + 2 k ) x + 4 = 0 ;
or x = 2 ( k + 2 ) ( 4 + 2 k ) ± ( 4 + 2 k ) 2 − 4 ( k + 2 ) ( 4 ) ;
or x = 1 ± 2 ( k + 2 ) 4 k 2 − 1 6 (i)
In order for the first parabola to always surmount the second, we require the discriminant in (i) to be non-positive so that no more than one real root is obtained (i.e. just a tangency point at x = 1 ). This can only occur for k ∈ [ − 2 , 2 ] . Hence, n = − 2 , m = 2 ⇒ m + n = 0 .