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When you have that you basically have: \begin{align*} \sqrt{x^2} & < \sqrt{1} \iff\\ |x| & < \pm 1 \end{align*} Doesn't that give you 2 options? $|x| < 1, |x| <-1$? The latter is false because $|x|$ is always positive right?

So to solve this $-1 < x < 1$. Correct?

N. F. Taussig
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SilenceOnTheWire
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    If I wrote $\sqrt{1^2}$ would you have $\pm 1$ or $|1|$? – Chinny84 Oct 13 '19 at 18:19
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    $|x|<1$ means that $\pm x < 1$, so $x<1$ and $-x<1$. Manipulating those gives you the result you state in your last line. – TM Gallagher Oct 13 '19 at 18:19
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    The square root function always returns the positive answer. There is a difference between $x^2=1$ and $x=\sqrt{1}$. – Michael Burr Oct 13 '19 at 18:20
  • What would “$|x| < \pm 1$” even mean? For equations, $x=\pm a$ means “$x=a$ or $x=-a$”, but for inequalities, putting “or” between the two cases leads to a rather strange statement (which is definitely not equivalent to the one you started with). – Hans Lundmark Oct 13 '19 at 19:04

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