this is about squaring both side for an absolute inequalities on only one side problem. For example:

$|6-2x|< x+4$ when solved both by squaring both sides and by defining it$ -(x+4)<6-2x< x+4$. they both give the same answer.

But for $|x+1|<2x+5$ squaring both side give out answer: $x<-4$ and $x>-2$ ;

by defining it $-(2x+5)< x+1<2x+5$, it give out answer: $x>-2$ only.

and when i check for $x<-4$ using $x=-5$, the solution for $x<-4$ contradict $|x+1|<2x+5$.

So, how do we know if the expression can be solved by squaring both sides? thank you


2 Answers 2


If you square both sides you may introduce extraneous roots. Therefore, if you do square both sides, you must check your answers and throw away those that do not work.

Working from the definion

$$|a|<b\iff (-b<a\le 0 \quad \lor \quad 0\le a<b )$$

you will not get extraneous roots. That may take longer, as in your second example, since you have four inequalities that combine in a non-trivial way thus you may get a contradiction or a redundancy. But that does not happen often, and only when the variable is in the $b$ expression.

So decide: do you want to avoid checking extraneous roots or handling four inequalities? You can always square both sides but that may not be the easiest way. I always go for the four inequalities.

  • $\begingroup$ if i stick to squaring both side and introduced extraneous roots. is there any possible restriction i could use to restrict the use of squaring both side? thank you. $\endgroup$
    – d_1
    Jul 1, 2015 at 1:44
  • $\begingroup$ @d_1: The only restriction I can think of is if both sides of the equation are clearly both positive (as in $\sqrt x=2$) or both negative (as in $-\sqrt x=-2$). $\endgroup$ Jul 1, 2015 at 10:14

In order to square both sides you need that $x+4 > 0$ or $x > -4$, then you can square now: $(6-2x)^2 < (x+4)^2 \iff 36-24x+4x^2 < x^2+8x+16 \iff 3x^2-32x+20 < 0 \iff (3x - 2)(x - 10) < 0 \iff \dfrac{2}{3} < x < 10$. Note that $x > \dfrac{2}{3} > -4$. So the solution is: $\left(\dfrac{2}{3}, 10\right)$


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