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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.

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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 $$ (-\infty,3) $$

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  • $\begingroup$ Your answer prevented me from making mistakes in my edit. Thanks. $\endgroup$ – drhab Jul 14 '15 at 12:17
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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:

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

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

  • $x<3$

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  • $\begingroup$ i understand that, thanks. but i need to show a canonical way. $\endgroup$ – R K Jul 14 '15 at 11:30
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The answer is (d), which is the union of the solutions for the two cases.

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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 $$\frac{5^2−5\cdot 5+12}{5−3} = \frac{25−25+12}{2} = \frac{12}{2}=6\geq 0$$.

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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 $x<3$.

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