# A question in inequalities involving modulus operator

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.

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

HINT:

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 $\, [-4,3) \,\cup\,[4,\infty)\,$.

• So, should I not consider that? – Wrichik Basu May 10 '17 at 18:47
• 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? – Wrichik Basu May 10 '17 at 18:50
• @WrichikBasu I edited the answer and expanded on the hint some more. – dxiv May 10 '17 at 19:02
• but overall I'm getting: $x\in(-\infty,-4]\cup[4,\infty)$; can you please check @dxiv whether I'm right or not? – k.Vijay May 10 '17 at 19:06
• @k.Vijay OP's answer is right, see the latest edit. – 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\$.