solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$. 
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.

options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
  $
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
 A: The equality is not true if $x=3$ so cases a),b) and c) can immediately be excluded.

Edit:
The following statements are equivalent:


*

*$\frac{x^{2}-\left|x\right|-12}{x-3}\geq2x$

*$\frac{x^{2}-6x+\left|x\right|+12}{x-3}\leq0$

*$\left[\frac{x^{2}-5x+12}{x-3}\leq0\wedge x\geq0\right]\vee\left[\frac{x^{2}-7x+12}{x-3}\leq0\wedge x<0\right]$

*$\left[\frac{x^{2}-5x+12}{x-3}\leq0\wedge x\geq0\right]\vee\left[\frac{\left(x-3\right)\left(x-4\right)}{x-3}\leq0\wedge x<0\right]$
We find a negative discriminant for $x^{2}-5x+12$ and conclude that
this expression is positive. Then we proceed with the following equivalent
statements:


*

*$\left[x-3<0\wedge x\geq0\right]\vee\left[x-4\leq0\wedge x\neq3\wedge x<0\right]$

*$\left[0\leq x<3\right]\vee\left[x<0\right]$

*$x<3$
A: You're correct in dividing between $x\ge0$ and $x<0$.
Case $x\ge0$: the inequality is
$$
\frac{x^2-x-12}{x-3}\ge 2x
$$
that becomes
$$
\frac{x^2-x-12-2x^2+6x}{x-3}\ge0
$$
or
$$
\frac{x^2-5x+12}{x-3}\le0
$$
Since the numerator is positive (as the discriminant is negative), the solution set for this inequality is the interval $[0,3)$ (that is, $0\le x<3$), not the empty set. (This is where you got wrong.)
Case $x<0$: the inequality is
$$
\frac{x^2+x-12}{x-3}\ge 2x
$$
that becomes
$$
\frac{x^2+x-12-2x^2+6x}{x-3}\ge0
$$
or
$$
\frac{x^2-7x+12}{x-3}\le0
$$
or
$$
\frac{(x-3)(x-4)}{x-3}\le0
$$
which is satisfied for $x\le 4$ and $x\ne3$, but we have also the condition $x<0$, so the solution set is $(-\infty,0)$.
Putting together the two cases, we get
$$
(-\infty,3)
$$
A: The answer is (d), which is the union of the solutions for the two cases.
A: Answer to question:
If we take $x=3$, we see that the equality doesn't hold, thus answers a, b and c are wrong. The only option left is d.
In case 1 you made an error, this equality doesn't hold:
$$\frac{x^2−5x+12}{x−3}≤0$$
 if you take $x=5$ we see that
$$\frac{5^2−5\cdot 5+12}{5−3} = \frac{25−25+12}{2} = \frac{12}{2}=6\geq 0$$.
A: Case $x>3$ : 
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2-x-12 \geq 2x(x-3) \Rightarrow x^2-x-12 \geq 2x^2-6x \\ \Rightarrow x^2-5x+12 \leq 0 $$ $\Delta=25-48<0 \text{ That means that } x^2-5x+12 \text{ has always the sign of } x^2 \\ \text{ so it is always positive. So we reject this case. } $ 
Case $x<3$ :  


*

*Subcase $0<x<3$ : 
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2-x-12 \leq 2x(x-3) \Rightarrow x^2-x-12 \leq 2x^2-6x \\ \Rightarrow x^2-5x+12 \geq 0 $$ $\Delta=25-48<0 \text{ That means that } x^2-5x+12 \text{ has always the sign of } x^2 \\ \text{ so it is always positive. So  } \checkmark $ 

*Subcase $x<0$ : 
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2+x-12 \leq 2x(x-3) \Rightarrow x^2+x-12 \leq 2x^2-6x \\ \Rightarrow x^2-7x+12 \geq 0 $$ $\Delta=49-48=1>0  \\ x_{1,2}=\frac{7 \pm 1}{2}= 3 \text{ or } 4 $ 
For $x \in (-\infty, 3] \cup [4, +\infty)$ we have that $x^2-7x+12 \geq 0$ but since we have that $x<0$ we conclude that in the case $x<0$ $x^2-7x+12 \geq 0$. 
In conclusion, it stands that $\frac{x^2-|x|-12}{x-3} \geq 2x$ for $x<3$. 
