Solve the inequality $\sqrt{\frac{4x-1}{x-a}}>a$ 
Solve the inequality
  $$\sqrt{\frac{4x-1}{x-a}}>a$$

My work so far:
If $a<0$ then $$\frac{4x-1}{x-a}\ge0$$
$$x\in(-\infty;a)\cup[\frac14;+\infty)$$
Let $a\ge0$ then 
1) $$\frac{4x-1}{x-a}\ge0$$
and 
2) $$\frac{4x-1}{x-a}>a^2$$
I need help here.
 A: if $a>0$, 
$\dfrac{4x-1}{x-a}$ has to be positive. so if $ a>x$ the top should be negative so :
$$4x-1<0$$
$$x<1/4$$
$$x\in (-\infty, u) $$
where u is minimum of ${a,1/4}$,
if $a< x$, so $x$ is positive and the top should be positive so:
$$4x-1>0$$
$$x>1/4$$
$$x\in (u,\infty) $$
wher u is maximum of ${1/4 , a}$
the final answer:
$$x\in (-\infty,v)\cup(w,\infty)$$
where $v$ is minimum of $a$ and $1/4$, and $w$ is maximum of $a$ and $1/4$
A: If $a \ge 0$, then we can square both sides.
$$\frac{4x-1}{x-a} > a^2$$
We need $x \ne a$.
Multiply both sides by $(x-a)^2$,
$$(4x-1)(x-a) > a^2(x-a)^2$$
$$(x-a)(4x-a^2x-1+a^3) >0$$
$$(x-a)((4-a^2)x-(1-a^3)) >0$$
To solve if further, we need to figure out when does $$a> \frac{1-a^3}{4-a^2}$$
Multiply $(4-a^2)^2$ on both sides, we have
$$a(4-a^2)^2 > (1-a^3)(4-a^2)$$
which is equivalent to 
$$(4-a^2)(4a-a^3-1+a^3)>0$$
$$(2-a)(2+a)(4a-1)>0$$
$$(a-2)(2+a)(4a-1)<0$$
$$(a-2)(4a-1)<0$$
$$\frac14 < a< 2$$
We consider $4$ cases: 
Case $1$： if $a \le \frac14$, then $x<a$ or $x > \frac{1-a^3}{4-a^2}$.
Case $2$: if $\frac14<a<2$, then $x < \frac{1-a^3}{4-a^2}$ or $x > a$.
Case $3$: if $a =2$, we have $x > 2$.
Case $4$: if $a > 2$, we have $$(x-a)((a^2-4)x-(a^3-1))<0$$
Hence $$a<x<\frac{1-a^3}{4-a^2}$$
