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I am doing a problem and am not sure why the final step is the following

$x^{2}<4 <=> -2<x<2$


When simplifying $x^{2}<4$ it becomes $x<±2$ which is the same as $x<2$, $x<-2$

but this does not match the actual inequality of $-2<x<2$, since $x>-2$ not $x<-2$ which is one of the inequalities obtained in the line above

My question:

Inequalities are swapped when multiplying & dividing by negative numbers, so is the reason why $x^{2}<4 <=> -2<x<2$

because square rooting both sides of $x^{2} <4$ give $x<2$ AND $x >-2$ since $4$ became $-2$ so it is like the inequality rule of dividing by a negative number? Hence the sign is flipped when considering $√(4)=-2$? $<=> 4/-2=-2$?


What I think:

$x^{2}<4$ gives four possible inequalities:

(1)$x<2$

(2)$-x<-2=>x>2$

(3)$-x<2=>x>-2$

(4)$x<-2$.

By inspection, (1) & (3) are the actual inequalities and (2) & (4) are false solutions.

(1) & (3) = $x<2$, $x>-2$ $<=>$ $-2<x<2$

Is this the reason why??

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  • $\begingroup$ Because the real numbers with $x^2<4$ are precisely those between $-2$ and $2$? $\endgroup$ Jul 26, 2020 at 6:32
  • $\begingroup$ I know one can easily see why $x^{2}<4 <=> -2<x<2$ but I want to know why the simplification does not match this inequality $\endgroup$
    – user314
    Jul 26, 2020 at 6:33
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    $\begingroup$ You wrote $x<-2$ is a solution but that is wrong. Take for example $x=-3$ then $x^2=9>4$. You must do the flip. $\endgroup$ Jul 26, 2020 at 6:54
  • $\begingroup$ Yes I know it is wrong. That is a result from simplifying the inequality which gives an incorrect inequality. I want to know why this is $\endgroup$
    – user314
    Jul 26, 2020 at 7:06

4 Answers 4

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Care must be exercised when considering negative square roots. For example, we have $9\gt4$, but not $-3\gt-2$.

So we can only take positive square roots.

From $x^2\lt4$ and $(-x)^2\lt4$ this gives two equations:

  • $x\lt2$
  • $-x\lt2\implies x\gt-2$
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  • $\begingroup$ Wait a second I think thats actually wrong $\endgroup$
    – user314
    Jul 26, 2020 at 6:44
  • $\begingroup$ $-x<-2==>x>2$ not $x>-2$ $\endgroup$
    – user314
    Jul 26, 2020 at 6:45
  • $\begingroup$ dividing $-1$ on both sides give $x>2$ not $x>-2$ My only guess why the inequality is $x>-2$ is because $√4 = 2,-2$ where $√4=-2 <=> 4/-2=-2$ so inequality sign is flipped. Not sure if this is correct $\endgroup$
    – user314
    Jul 26, 2020 at 6:52
  • $\begingroup$ Not entirely sure if this matches your answer but I noticed that $x^{2}<4$ gives four possible inequalities: (1)$x<2$, (2)$-x<-2=>x>2$, (3)$-x<2=>x>-2$ (4)$x<-2$. By inspection, (1) & (3) are the actual inequalities and (2) & (4) are false solutions. $\endgroup$
    – user314
    Jul 26, 2020 at 7:04
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Your thinking is correct but it is quite tedious to consider the four possible inequalities.

A faster method would be simply observing the fact that,

$$|x|<2$$ where $x^{2}<4$ is true simply when the magnitude of $x$ is less than $2$ (e.g. $x=0,1,1.3,1.999$)

Note that |x| is a piecewise function $$|x| = \begin{cases} -x\\ x \end{cases}$$ Therefore, we obtain

  • $x<2$
  • $-x<2 \Rightarrow x>-2$

$$-2<x<2$$

Also if you were to simplify the inequality using algebra like you have shown in your question, you would have to look through the four inequalities and see which are valid. As mentioned that is tedious, so you should just consider the magnitude of $x$ and make use of $|x|$

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You have to deal with inequalities carefully:

$x^2 < 4 \iff x^2-4<0 \iff(x-2)(x+2)<0 \iff -2< x < 2$

Now can you see why?

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  • $\begingroup$ That is still the same as square rooting 4 since solving $(x-2)$ and $(x+2)$ individually give the inequalities, respectively $x<2$ and $x<-2$ which still contradicts $-2<x<2$ $\endgroup$
    – user314
    Jul 26, 2020 at 6:36
  • $\begingroup$ Regardless of the approach to this question, does the sign of $x<-2$ get flipped because 4 turning into -2 kind of implies division by a negative number? The result would then be x > -2, which is correct $\endgroup$
    – user314
    Jul 26, 2020 at 6:38
  • $\begingroup$ No it isn't. Try to figure out what values of $x$ can cause $(x-2)(x+2) < 0$. $\endgroup$ Jul 26, 2020 at 6:38
  • $\begingroup$ Simply looking at it, the answer would be $-2<x<2$ but simplifying the inequality $(x-2)(x+2)<0$ contradicts $-2<x<2$. Dividing $(x+2)$ on both sides gives the inequality $x<2$ which is correct. But dividing $(x-2)$ on both sides gives the inequality $x<-2$ which contradicts the actual inequality of $x>-2$ $\endgroup$
    – user314
    Jul 26, 2020 at 6:59
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You must take care when simplifying the inequality:

You say that $(1):x<2 , (2): x<−2$

Squaring $(2)$: $x^2>4$. The reason this is, to square $-2$ is to multiply by a negative on the RHS (-2). And on the LHS, you are also multiplying by a negative since $x$ is less than -2 (hence negative).

Squaring $(1)$: $x^2<4$. Well that contradicts $(2)$. So you have made an error.

Also: $x<2$ and $x<-2$ implies $x<-2$, and I showed previously that this will imply that $x^2>4$, which contradicts your premise.

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