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Suppose we have $$ x^{2} = (-x)^{2}.$$

I understand that this equation holds because $$\begin{aligned} (-x)^{2} & = (-1\cdot x)^{2} \\ & = (-1)^{2} \cdot x^{2} \\ & = x^{2}, \end{aligned}$$ and so $\sqrt{x^{2}} = \sqrt{x^{2}}$ becomes $x = x$ and everything is fine.

But what if we just take the square root first, like so: $$\sqrt{x^{2}} = \sqrt{(-x)^{2}}.$$ Wouldn't this gives $x = -x$, which is a contradiction?

Tan2525
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    No. The answer is not $x=-x$ but $|x|=|-x|$. By definition, $\sqrt{\ }:[0,\infty)\to[0,\infty)$ – Andrei Feb 28 '21 at 15:19
  • $\sqrt{a}$ where $a$ is a positive real number gives the unique positive real number $b$ such that $b^2=a$. For instance, $\sqrt{16}=4$ and only $4$. Here, $\sqrt{~}$ is being used as a function which means that it only has one output. It is not true that $\sqrt{x^2}=x$ unless we know that $x$ were positive. Instead $\sqrt{x^2}=|x|$ – JMoravitz Feb 28 '21 at 15:22
  • It seems that I forgot about that part of the definition of square root. Thank you @Andrei and JMoravitz. – Tan2525 Feb 28 '21 at 15:25
  • @JMoravitz Yes, my question has been resolved. Thanks. If someone wishes to submit an answer I'll mark it accordingly. – Tan2525 Feb 28 '21 at 15:26

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Since $\sqrt{x}$ denotes the nonnegative square root of $x$, it is not true in general that $\sqrt{x^2}=x$. For instance, if $x=-5$, then $$ \sqrt{x^2}=\sqrt{25}=5=-x \, . $$ So sometimes $\sqrt{x^2}$ equals $x$ and sometimes it equals $-x$, depending on the sign of $x$. In general, $$ \sqrt{x^2} = \begin{cases} x \text{ if $x\geq0$} \\ -x \text{ if $x\leq 0$} \end{cases} \, . $$ What makes things confusing is that if $x<0$, then $-x$ is positive.

Joe
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