Help required in this inequality:

If $$\frac {x^2-|x|-12}{x-3} \ge 0 $$ prove that $x \in [-4,3) \cup [4,\infty) \quad $.

Problem I'm facing:

I know that $x^2$ can be made $|x|^2$, but what about the $x $ in denominator? If that can be changed to $|x|$, then I can easily solve by wavy curve method.

N.B.: I'm a beginner in pure maths and will perhaps remain so for the rest of my life. So, please explain simply and don't complicate things.

  • $\begingroup$ but I'm getting: $x\in(-\infty,-4]\cup[4,\infty)$ $\endgroup$ – k.Vijay May 10 '17 at 19:04


For Wavy Curve method always try to find zeroes of the equation(numerator and denominator separately) and then plot them on number lines.

from numerator, the zeroes are $4$ and $-4$. and from denominator $3$.

You cant take $3$ as the function will not be defined on $3$.

Now construct a number line and check for what values of $x$ function is positive.That would be the desired result.


Hint:  write it as $\;\cfrac{(\,|x|-4\,)(\,|x|+3\,)}{x-3} \ge 0\,$ and check the signs of each factor. For example, $\,|x|+3 \gt 0\,$ for $\forall x \in \mathbb{R}$ which leaves only two factors to consider.

[ EDIT ] This leaves two possibilities to consider:

  • $|x|-4 \ge 0$ and $x-3 \gt 0\,$, or

  • $|x|-4 \le 0$ and $x-3 \lt 0\,$

The first case, for example, implies $|x| \ge 4$ and $x \gt 3\,$, giving:

$$x \in \big( (-\infty,-4] \cup [4,\infty) \big) \;\cap\; (3,\infty) \,=\, [4,\infty)\,$$

The second case can be worked out similarly for $|x| \le 4$ and $x \lt 3\,$, giving:

$$x \in [-4,4] \;\cap\; (-\infty, 3) \,=\, [-4,3)\,$$

Putting the two together, the final answer is $\, [-4,3) \,\cup\,[4,\infty)\,$.

  • $\begingroup$ So, should I not consider that? $\endgroup$ – Wrichik Basu May 10 '17 at 18:47
  • $\begingroup$ Then $x-3>0$ as denominator cannot be 0. It leaves me with $|x|-4 \ge 0$, which means $|x| \ge 4$, so $x =-4,4. $ Then? $\endgroup$ – Wrichik Basu May 10 '17 at 18:50
  • $\begingroup$ @WrichikBasu I edited the answer and expanded on the hint some more. $\endgroup$ – dxiv May 10 '17 at 19:02
  • $\begingroup$ but overall I'm getting: $x\in(-\infty,-4]\cup[4,\infty)$; can you please check @dxiv whether I'm right or not? $\endgroup$ – k.Vijay May 10 '17 at 19:06
  • $\begingroup$ @k.Vijay OP's answer is right, see the latest edit. $\endgroup$ – dxiv May 10 '17 at 19:14

Taking hint from @Dynamo, I did it in this way:

For finding critical points:

Factorising, numerator = $(|x|-4)(|x|+3) $.

Now, $|x|+3=0$ means $|x| =-3, which is meaningless and hence discarded.

$|x|-4 $ has two critical points for $x $, viz. 4 & -4, both closed. Denominator has critical point 3, open.

Now I've put it on the curve and solved. Thanks @dynamo.


Solve the equation on three intervals:

  • $x \leq 0$; here $|x| = -x$ and $x-3 < 0$,
  • $0 \leq x \leq 3$; here $|x| = x$ and $x-3 \leq 0$,
  • $x \geq 3$; here $|x| = x$ and $x-3 \geq 0$.

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