Square root of a negative number squared

Question

√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

Answer 1

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.

Answer 2

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 $\sqrt{25}=5$.

Answer 3

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 $5i$.

Answer 4

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.