# what is the square root of $-4$?

What is the square root of $-4$ (negative 4)?

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You could accept an answer by clicking the check mark on whichever one works best for you. If none do, please improve the question in a comment. Thanks. –  Ross Millikan Dec 11 '10 at 5:14
@Ross: I'd say that at the moment it's quite difficult to choose an answer: Stictly mathematically speaking, only Yuval's answer is correct; the problem with the other answers is explained a bit here –  Hendrik Vogt Dec 11 '10 at 8:47
Why the downvoting of the answers? –  user02138 Dec 16 '11 at 4:14

The main point should be find out the square roots of -1.In complex number system,it is known that the roots of the equation $x^2+1=0$ is ±i. "$i$" is the unit of imaginary number.So we get $(±i)^2=-1$,which means the square roots of -$1$ are $±i$.Then you can factor -4 to $(-1)*4$，and calculate its square roots which should be $±2i$.
Actually,not only for 4,the method to factor a negative number $-n$ ($n$ is an arbitrary positive integer) to $(-1)*n$ can also help to work out its square roots which are $±\sqrt{n}i$ in the complex number system.

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Recall that $\sqrt{4} = \pm 2$. Since $-4 = (4)(-1)$, we have $\sqrt{-4} = \sqrt{4} \sqrt{-1} = \pm 2 i$, where $i = \sqrt{-1}$. Note that these are the two roots of the quadratic $x^{2} + 4 = 0$, which has no real roots (positive or negative) by the Descartes Rule of Signs. This last statement rules out the case that the root is $-2$, which is a common mistake for newcomers to elementary algebra.

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Alternatively, if one didn't remember Descartes's rule of signs, there are no real roots because $x^2+4\geq 4$ when $x$ is real. –  Jonas Meyer Dec 11 '10 at 5:06
Why the downvote? –  user02138 Dec 16 '11 at 4:11
No problem. Thanks. –  user02138 Dec 16 '11 at 4:30

When you say the square root of $-4$, rather than a square root, the notion of "principal square root" comes to mind.

So while it is true that $-2i$ and $+2i$ are each $a$ square root of $-4$, it is convention to choose $i \cdot \sqrt x$ to be the square root of $-x$ (for $x > 0$ )

The introduction of the equation $x^2 = -4$ and its solutions ($x = \pm 2i$ ) confuses the issue.

In summary, $-2i$ is a square root of $-4$.

$+2i$ is the square root of $-4$.

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no it's not -2.

The square root is a number $x$ such that its square is the original number. Depending on the number system you are looking at there might be no square root, for example in the integers or reals. Over the complex numbers there is a square root which other posters have given as 2i or -2i

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There is no solution to this in the real numbers. The square of any number is greater or equal to zero.

In complex numbers however, we have defined something called $i$ with $i \cdot i = -1$ or equivalently $i = \sqrt{-1}$

Thus

$$x^2 = -4 = 4\cdot(-1)$$ $$x = \pm \sqrt{4}\cdot\sqrt{-1} = \pm2i$$

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I wouldn't say "equivalently $i = \sqrt{-1}$". This last sentence is really not equivalent to $i^2 = 1$, because the writing $i = \sqrt{-1}$ suggests there is only one square root of one, where there are actually two. Using this notation can lead to contradictions ; it must only be used when vaguely speaking. –  Patrick Da Silva Nov 24 '12 at 5:02

-4 has no real square roots, since $x^2 \geq 0$.

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so simply an error because -2 * -2 is 4 and not -4 so it give you an error? –  carefacerz Nov 29 '10 at 17:42
@Haujpaw: your calculator might say "error". Mathematicians like Argand and Gauss have tried to find ways to solve equations that involve 'calculating' the root of a negative number. With the help of these mathematicians, we can now use the Complex number set to solve such equations (see wikipedia). Like Dario mentioned, we simply define $i=\sqrt(-1)$. Using this definition, we see that $x^2=-4$ gives x is plus or minus 2i, just like $x^2=4$ gives plus or minus 2. –  Max Muller Nov 29 '10 at 20:08
sorry i dont see hot you combine x and i. where do you use i? i know you define it but you didnt include it when : x2=−4 gives x is plus or minus 2i, just like x2=4 gives plus or minus 2. –  carefacerz Nov 29 '10 at 21:08
@ carefacerz: I don't understand what you're trying to ask. If you want to find out more about complex numbers in general, I suggest you to take a look at amazon.com/Complex-Numbers-Z-Titu-Andreescu/dp/0817643265/… or go to university :)! –  Max Muller Nov 30 '10 at 16:02