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

absolute valuealgebra-precalculusinequalityquadratics

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


$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.


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:

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

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

  • $x<3$

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