8√[4x + 2] > 5122x / 8√[16x + 8]
8√[4x + 2] > 5122x / 8√[4(4x + 2)]
8√[4x + 2] > 5122x / 8 2√[(4x + 2)]
8√[4x + 2] > (83)2x / 8 2√[(4x + 2)]
8√[4x + 2] > 8 6x - 2√[(4x + 2)] we can solve for the exponents.......
√[(4x + 2)] > 6x - 2√[(4x + 2)]
√[(4x + 2)] > 6x - 2√[(4x + 2)] add 2√[(4x + 2)] to both sides
3 √[(4x + 2)] > 6x divide both sides by 3
√[(4x + 2)] > 2x square both sides
4x + 2 > 4x^2 divide through by 2
2x + 1 > 2x^2
0 > 2x^2 - 2x - 1
And the solution to this is 1/2 (1-sqrt(3)) < x <1/2 (1+sqrt(3))
However, for the original equation √[(4x + 2)] > 2x ......this is true if -1/2 ≤ x <1/2 (1+sqrt(3))
But, since 1/2 (1-sqrt(3)) > -1/2 , we can expand the solution interval to -1/2 ≤ x < 1/2 (1+sqrt(3))