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√x^2=|x|,
What about √-x^2 ?
If we use the number $5$ as en example, would this evaluate to √-5^2 = √25 =5
OR do we need to get the imaginary number 'i' involved, resulting in √-5^2 = 5i
I have found many conflicting answers
Any clarification on the topic would be much appreciated

Kind Regards

mvw
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Amanda
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    I have no idea what you want to know. – Suzu Hirose Nov 26 '14 at 10:32
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    Note that $-5^2\neq 25$ but rather $-25$. But the main thing is that the symbol $\sqrt{}$ is only well-defined to take a single value (in a nice way) when the argument is a non-negative number. – Tobias Kildetoft Nov 26 '14 at 10:34
  • There is a convention in mathematics that $-5^2$ is $-25$, not $25$. The square is evaluated before the minus. The square root of $-25$ can be defined to be $5i$ or $-5i$. – Peter Franek Nov 26 '14 at 10:38
  • Thank you peter, your answer is perfect. – Amanda Nov 26 '14 at 10:45
  • Amanda, can you add parentheses to make clear which part of the term is squared and which part is rooted? Now it is ambigious. – mvw Nov 26 '14 at 10:48
  • Caution, your title does not match the question and your example $\sqrt{-5^2}$ gives $\sqrt{-25}$, not $\sqrt{25}$. –  Nov 26 '14 at 10:53

4 Answers4

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In real numbers, the square root of a negative is not defined.

$$\color{red}{\sqrt{-25}}.$$

In complex numbers, the square root of a negative can indeed be defined as $i$ times the square root of the absolute value.

$$\sqrt{-25}=5i.$$

But you need to understand the concept of principal branch, as $-5i$ could also be an acceptable answer.

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    +1 for the link to the principal branch. It is so easy to simply say that $\sqrt{-1}=i$ without mentioning that expanding square roots to complex numbers is not trivial. – 5xum Nov 26 '14 at 12:00
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There is a HUGE difference between

$$(\sqrt{-5})^2$$

which is not a well defined expression because the square root is a function only defined on the positive real numbers (and can be expanded to the whole complex numbers, but it is not trivial and there actually exist multiple expansions of the function, since both $i^2$ and $(-i)^2$ evaluate to $-1$)

and $$\sqrt{(-5)^2}$$

which is equal to $5$ since $\sqrt{25}=5$.

5xum
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The other answers currently written do not hit on what I believe to be the true source of confusion here. The ambiguity in the answer lies not with the square root... but rather on the negative sign.

The value $-5^2$ is most commonly interpreted as $-(5^2)$ which simplifies as $-25$. It is distinctly different than the value $(-5)^2$ which simplifies as $25$.

The result of $\sqrt{(-5)^2}$ is indeed positive $5$.

The result of $\sqrt{-(5^2)}$ is $5i$.

JMoravitz
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The answer is ±5.

$-5$ is converted to $5i^2$ then squared, resulting to $25i^4$, according to the PEMDAS rule.

The root of $25i^4$ is then evaluated by taking the roots of a complex number. Note that $z = r(\cos\phi + i\sin\phi)$, where $z$ is the $n$-th root.

$r$ is the real root part, but in this case is used to contain imaginary numbers as well. Therefore, $r = \sqrt{25i^4} = 5i^2$, without regard to sign.

$\phi = 360°/n$, where $n$ is the $n$-th root. Thus, $n = 2$, therefore $\phi = 180°$.

Using the argand diagram, there are two cases, $\phi = 0°$ and $\phi = 180°$. Evaluating $z$, we have $z = -5i^2$ ($\phi = 180°$) and $z = 5i^2$ ($\phi = 0°$), giving us $z = ±5$ upon simplification.

The only thing I am not so sure about is if it is allowed to use imaginary numbers in $r$, as this is counter-intuitive from its derivation, and I'm gonna confirm it with my math doctor.

Hope this helps.

Glorfindel
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Kurt
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