What is the difference between $\sqrt{x^2}$ and $(\sqrt{x})^2$? [closed]

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$\sqrt{x^2}\;$ and $(\sqrt{x})^2$; I mean, are they equal? because what about the following facts:$\sqrt{x^2} = (x^2)^{1/2} = x^{(2)(\frac{1}{2})} = x^{\frac{2}{2}} = x \\ (\sqrt{x})^2 = x^{\frac{1}{2}(2)}= x^{(\frac{1}{2})(2)} = x^{\frac{2}{2}} = x$

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2 Answers

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As functions over the reals,
the domain of $\sqrt {x^2}$ is $\mathbb R$,
the domain of ${\sqrt x}^2$ is $[0,\infty)$.

Verbally, in the same order,
the square root of (x squared),
the square of (square root of x).

Does that clarify the order?

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The point is that in the first you square and then take the square root. That works well even if $x$ is negative. The second has you take the square root first, which you can't do if $x$ is negative, then square the result. As long as $x$ is positive the two are equal, but if $x$ is negative the second is undefined. The point of the question is the order of operations.

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