I am confused about how square roots work.
I know the square root function gives the principal square root, but I'm confused because if we just take the positive number, I get $x = 4$, which doesn't include the $x=-4$ case.
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3$\sqrt{x^2}=|x|$ – Geoffrey Trang Jul 20 '21 at 20:34
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Have you tried plugging in $x=-4$ to the equation $\sqrt{x^2}=\sqrt{16}$ to see if equality holds? – Joe Jul 20 '21 at 20:35
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Unless I'm crazy I'm quite certain the equality holds: $\sqrt{(-4)^2} = \sqrt{16}$ – beginner Jul 20 '21 at 20:36
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So I should view the answer as $\vert x \vert = 4$? – beginner Jul 20 '21 at 20:39
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2@beginner: Yes, because $(-4)^2$ is the same number as $16$, and $\sqrt{\cdot}$ is a function that maps every nonnegative real number to its nonnegative real root. The number $16$ has two square roots, $4$ and $-4$, but only one of these is nonnegative. And the solution to your overall equation is indeed $|x|=4$, which is equivalent to $x=\pm 4$. – Joe Jul 20 '21 at 20:41
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Thank you very much! – beginner Jul 20 '21 at 20:48
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1@beginner: No problem! – Joe Jul 20 '21 at 22:08