Answer:
Below in bold.
Explanation:
One way to do these is to square both sides:
(x + 4)^2 < 4x^2
x^2 + 8x + 16 < 4x^2
3x^2 - 8x - 16 > 0
Let this = 0:
3x^2 - 8x - 16 = 0
(3x + 4)(x - 4) = 0
x = -4/3, 4.
So the critical points are -4/3 and 4.
Make a table:
x < - 4/3 -4/3 +< x <= 4 x > 4
3x + 4 <0 > 0 > 0
x - 4 <0 < 0 > 0
(3x+4)(x-4) >0 <0 > 0
So the solution is x < -4/3 or x > 4
or in interval notation:
(-∞, -4/3) U (4, ∞)