How do you simplify ${\sqrt{3}}^{2}$ $sqrt3^2$ ?

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How do you simplify ${\sqrt{3}}^{2}$ $sqrt3^2$ ?

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Answer 1 · 2016-02-03T19:48:42

Ambiguous

Explanation:

I cannot understand if the square is inside or outside the radix. Anyhow, if it's inside (the most plausible one...)
$\sqrt{3^{2}}$ $sqrt(3^2)$ it's one of the two numbers $\pm a$ $+-a$ such that ${(\pm a)}^{2} = 3^{2} = 9$ $(+-a)^2=3^2=9$ , so they are $\sqrt{3^{2}} = \pm 3$ $sqrt(3^2)= +-3$
On the other hand, if we consider ${(\sqrt{3})}^{2}$ $(sqrt(3))^2$ , you gotta take both $\pm \sqrt{3}$ $+-sqrt(3)$ and take the square. But, by definition, $\pm \sqrt{3}$ $+-sqrt(3)$ are the only two numbers such that ${(\pm \sqrt{3})}^{2} = 3$ $(+-sqrt(3))^2=3$ , so the answer is $3$ $3$

Answer 2 · 2016-02-03T19:48:59

Ambiguous

Explanation:

I cannot understand if the square is inside or outside the radix. Anyhow, if it's inside (the most plausible one...)
$\sqrt{3^{2}}$ $sqrt(3^2)$ it's one of the two numbers $\pm a$ $+-a$ such that ${(\pm a)}^{2} = 3^{2} = 9$ $(+-a)^2=3^2=9$ , so they are $\sqrt{3^{2}} = \pm 3$ $sqrt(3^2)= +-3$
On the other hand, if we consider ${(\sqrt{3})}^{2}$ $(sqrt(3))^2$ , you gotta take both $\pm \sqrt{3}$ $+-sqrt(3)$ and take the square. But, by definition, $\pm \sqrt{3}$ $+-sqrt(3)$ are the only two numbers such that ${(\pm \sqrt{3})}^{2} = 3$ $(+-sqrt(3))^2=3$ , so the answer is $3$ $3$

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How do you simplify ${\sqrt{3}}^{2}$ $sqrt3^2$ ?

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