The quadratic equation $x^2+4mx+m = 2x - 6$ has exactly one real root. Find the positive value of $m$.
Simplify as
x^2 + (4m - 2)x + (m + 6) = 0
We have one root when the discriminant = 0
(4m - 2)^2 - 4 (1) ( m + 6) = 0
16m^2 - 16m + 4 - 4m - 24 = 0
16m^2 -20m - 20 = 0
4m^2 - 5m - 5 = 0
m = [ 5 + sqrt [ 25 + 80 ] ] / 8 = [ 5 + sqrt 105 ] / 8
+506 
