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.

  • $\begingroup$ It completely depends on the problem and squaring may be useful to get rid of lots of square roots, et cetera $\endgroup$ – user665856 Apr 25 '19 at 4:53
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    $\begingroup$ Is this $$\left|\frac{2x+1}{x-3}\right|\geq 2$$ ? $\endgroup$ – Dr. Sonnhard Graubner Apr 25 '19 at 5:33
  • $\begingroup$ Can you see why $|x|\ge2$ is the same thing as $x^2\ge4$? $\endgroup$ – Gerry Myerson Apr 25 '19 at 5:51
  • $\begingroup$ @Dr.SonnhardGraubner. No, he wrote 2x + 1/x - 3 >= 2. $\endgroup$ – William Elliot Apr 25 '19 at 6:11
  • $\begingroup$ @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. $\endgroup$ – Gregory Nisbet 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 $$-(2x+1)>-2(x-3)$$

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  • $\begingroup$ 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? $\endgroup$ – Andrew Lee 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 ?

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  • $\begingroup$ It is $x^2-6x+9$. $\endgroup$ – Word Shallow Apr 25 '19 at 11:02
  • $\begingroup$ Ooops ! Thanks. $\endgroup$ – Fred Apr 25 '19 at 11:03
  • $\begingroup$ 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 $\endgroup$ – Andrew Lee 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.

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  • $\begingroup$ 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? $\endgroup$ – Andrew Lee Apr 26 '19 at 6:26
  • $\begingroup$ @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 $E(x,y,\ldots).$ $\endgroup$ – Allawonder Apr 26 '19 at 10:57
  • $\begingroup$ @AndrewLee I have added some more explanation in my answer. See if it now makes some sense to you. $\endgroup$ – Allawonder Apr 26 '19 at 11:16

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