How is this statement true? "For any real number $x$ we have $\sqrt{x^2} = |x|$"? Because putting $x=2$ $\sqrt{x^2}$ gives BOTH $2$ and $-2$ But $|x|$ only gives $2$
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3Indeed $\sqrt{x^2}=|x|$,can be confuse as you said.... As convention, the square root function is only the positive part, so the statement is true. – Brethlosze Sep 18 '17 at 10:21
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Do you mean that by convention, √4 is only 2 and not -2? – user167573 Sep 18 '17 at 10:27
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Square root a number always gives a positive answer. – abc... Sep 18 '17 at 10:27
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2By convention, the square root function always gives a positive number, though both signs are valid. Hence $\sqrt{x}>0$ always. – Brethlosze Sep 18 '17 at 10:31
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4Your assertion that $\sqrt{4} = \pm 2$ is false. If $x$ is a real number, the notation $\sqrt{x}$ means the principal (nonnegative) square root of $x$. Also, see this related question. – N. F. Taussig Sep 18 '17 at 12:09
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Remember that by definition and for $x \in \mathbb{R}$ $$|x| = \begin{cases} x, & \mbox{if } x \ge 0 \\ -x, & \mbox{if } x < 0. \end{cases}$$
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Remember that the square root of $y\ge 0$ is the $x\ge 0$ such that $x^2=y$, not only the $x\in\mathbb{R}$ such that $x^2=y$.
For example:
$$\sqrt{3^2}=\sqrt{9}=3=|3|,$$
but
$$\sqrt{(-3)^2}=\sqrt{9}=3=|-3|.$$
MattAllegro
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