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

Question

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.

Answer 1

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 $$a<x<\frac{1-a^3}{4-a^2}$$

Answer 2

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 $1/4$