What is the square root of $-4$ (negative 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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-4 has no real square roots, since $x^2 \geq 0$. |
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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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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 $±2i$.
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Plus or minus 2i |
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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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protected by Zev Chonoles♦ Dec 16 '11 at 8:39
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