# Absolute Value Rational Inequalities Help Please

I have been read plenty of questions on questions like this, but I still dont quite get it. For example, this question:

$$\left| \frac{2x+1}{x-3} \right| \ge 2$$

How would I go about solving this?

-The method in the example went on to just square both sides of the equation and from there it formed a quadratic equation to solve it. But, I dont get why you can just square it, dont we need to take into consideration since the expression is absolute value, it could also have a negative value as well? Like, 2x+1 can also be -(2x+1). Is it because, regardless if it were positive or negative, once you square it, the result would be the same? I am confused because other times, typically we would have 2 cases, one for 2x+1 > 0 and 2x+1 <0.

Another questions is, when can I just square both sides and continue from there? Is there a more complete way of doing it rather than just squaring both sides and be done? In what scenarios, cant I square both sides? And what is the alternative method to solving those types of problems? Thanks.

• It completely depends on the problem and squaring may be useful to get rid of lots of square roots, et cetera
– user665856
Apr 25 '19 at 4:53
• Is this $$\left|\frac{2x+1}{x-3}\right|\geq 2$$ ? Apr 25 '19 at 5:33
• Can you see why $|x|\ge2$ is the same thing as $x^2\ge4$? Apr 25 '19 at 5:51
• @Dr.SonnhardGraubner. No, he wrote 2x + 1/x - 3 >= 2. Apr 25 '19 at 6:11
• @WilliamElliot ... Was / in the original ASCII inequality meant to have high or low precedence? I interpreted it as low precedence. If that was a mistake, someone please roll back my edit. Apr 25 '19 at 6:14

Hint: For $$x\neq 3$$ we can write $$|2x+1|\geq 2|x-3|$$ so we have to distinguish the following cases:

a) $$x>3$$ then we have $$2x+1\geq 2(x-3)$$ b)$$-\frac{1}{2}\le x<3$$ and we get $$2x+1>-2(x+3)$$ c) $$x<-\frac{1}{2}$$ and we have $$-(2x+1)>-2(x-3)$$

• Thanks, I have used this method. But I have a problem with the squaring both sides of the inequality method. I don't get how you can square |2x+1| >= 2|x-3| on both sides, knowing that if x were negative, the inequality sign would have to change? I watched a video proving why you can square both sides of an inequality. For example |a|^2 > |b|^2, then what he wrote next was (a)^2 > (b)^2. You can square the modulus because absolute value is always positive but if you get rid of the modulus, a could be negative and upon squaring a, the inequality symbol wouldould have to swap, right? Apr 26 '19 at 6:31

For $$x \ne 3$$ we have

$$\left|\frac{2x+1}{x-3}\right|\geq 2 \iff 4x^2+4x+1 \ge 4(x^2-6x+9)$$

Can you proceed ?

• It is $x^2-6x+9$. Apr 25 '19 at 11:02
• Ooops ! Thanks.
– Fred
Apr 25 '19 at 11:03
• Thanks! I get that you can square it, but how do you know that the expressions inside the modulus are non-negative? If they were, then the inequality symbol would have to change right? That's my problem here Apr 26 '19 at 6:21

You can square both sides of an inequality provided they are both nonnegative; this is because the squaring operation preserves order for nonnegative quantities. That is, given $$x,y\geq 0,$$ and $$x\leq y,$$ then it follows that $$x^2\leq y^2.$$ The reason is simple: Since both $$x$$ and $$y$$ are nonnegative, you can multiply $$x\leq y$$ by each in turn to get, respectively, $$x^2\leq xy$$ and $$xy\leq y^2,$$ from which the result follows immediately. Thus, you can square both sides of $$x\leq y$$ without qualms, provided that $$x,y$$ are nonnegative.

Recall that when the symbol $$|{\cdot}|$$ encloses an expression $$E(x,y),$$ say, we mean that the result -- here $$\left |E(x,y)\right |$$ -- is nonnegative, by definition -- that is, it is either positive, or else it vanishes. It follows from the explanation above that you can square both sides of your inequality, since $$\text{LHS}\ge 0,$$ and obviously $$2>0.$$

The explanation above holds, strictly speaking, only for inequalities involving $$\leq$$ (and of course, $$\geq$$), called weak inequalities; not for those involving the strong $$\lt.$$ The point of difference may appear minimal, but is essential from a strict perspective. If we have two nonnegative $$x,y$$ satisfying $$x\lt y,$$ then we can square them without disturbing the order $$\lt$$ only provided $$x,y$$ do not vanish simultaneously, for it is easy to see in that case that we have a false statement right from the beginning, namely $$0\lt 0.$$ This is the caveat to the above statements.

PS. This is for completeness, as OP seems to need more clarification in this direction.

I shall now solve your inequality in order to ascertain that everything I said is indeed clear. So we have $$\left| \frac{2x+1}{x-3} \right| \ge 2,$$ upon squaring both sides of which (this is legit as explained above) gives $$\left| \frac{2x+1}{x-3} \right|^2 \ge 4.$$

We pause here and note that the equality $$|A||B|=|AB|,$$ for any quantities $$A,B,$$ holds. This is a property satisfied by the modulus operation, whose proof would take us away from the main goal (try to convince yourself of its truth; by cases, perhaps). In that equality, if $$A=B,$$ then we have the true statement $$|A||A|=|A|^2=|AA|=|A^2|=A^2,$$ the last equality following provided that $$A$$ is real.

Now returning to the main thread and applying the above result, we then have $$\left| \frac{2x+1}{x-3} \right|^2= \left| \left (\frac{2x+1}{x-3} \right)^2\right|=\left (\frac{2x+1}{x-3} \right)^2\ge 4.$$

I believe you can now proceed from here.

• Thanks, but for here, how do we know if the expression inside the modulus is non-negative? For example, if the x were -2, and you were to cross multiply ending up with |2x+1| >= 2|x-3|. How can you square, both sides not knowing what the value of x is in the modulus? If x were -2 the inequality sign has to change right? Apr 26 '19 at 6:26
• @AndrewLee We don't care what's inside the moldulus symbol. We're squaring the modulus of whatever it is; since moduli are never nonnegative, it is legit if the other side of the inequality is also nonnegative. Thus, what you're squaring is $\left |E(x,y,\ldots)\right |,$ and not $E(x,y,\ldots).$ Apr 26 '19 at 10:57
• @AndrewLee I have added some more explanation in my answer. See if it now makes some sense to you. Apr 26 '19 at 11:16