You can re-write this quotient this way :
[tex]\frac{\sqrt{28(x-1)}}{\sqrt{8x^{2} }} = \sqrt{ \frac{28(x-1)}{8x^{2} } }\\[/tex]
What's inside the square root must be greater than or equal to 0, because the domain of the square root function is defined on R+ (which is [0,+∞))
In other word we must find x so that :
[tex]\frac{28(x-1)}{8x^{2}}\geq 0[/tex]
at the end we get
28(x-1) ≥ 0
28x - 28 ≥ 0
28x ≥ 28
x ≥ 1
So the answer is B, x ≥ 1
