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

absolute valuealgebra-precalculusinequalityquadratics

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.

#### Best Answer

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:

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

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

• $x<3$