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Possible Duplicate:
Square roots — positive and negative

$\sqrt{4} = -2$. WolframAlpha says "false"!

Now lets take a deeper look to my idea.

Well...we know that,

$$2^2 = 4 \iff \sqrt{4} = 2$$

$(-2)^2 = 4$ so why can't $\sqrt{4}$ be equal to $-2$?

I'm a little bit confused

// Thank you for all your answers, I have answer now. Stepo

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Stepo
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  • There are indeed two numbers that square to 4 but we have to choose which we meen by the square root. We usually choose the positive one. – fretty Oct 30 '12 at 18:05
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    $-2$ is a square root of four. $2$ is the square root of four (which is what that symbol means). – The Chaz 2.0 Oct 30 '12 at 18:05
  • @user2838619, the [tag:roots] tag isn't relevant for questions about square roots or other radicals. Use [tag:radicals] instead. – Antonio Vargas Jul 30 '14 at 23:01
  • Just a quick supplement to the answers you got - it may help to think about $\sqrt{x}$ as the side length of a square with area $x$. – Supware Feb 12 '18 at 16:49

3 Answers3

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It's just notation most likely. Yes, $(-2)^2 = 4$, but often the $\sqrt{4}$ symbol is reserved for the positive square root, so $\sqrt{4} = 2$. If you want the negative square root, that would be $-\sqrt{4} = -2$. Both $-2$ and $2$ are square roots of $4$, but the notation $\sqrt{4}$ corresponds to only the positive square root.

GeoffDS
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By definition, $$\sqrt{x^2} = \vert x \vert$$ for $x \in \mathbb{R}$.

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    That's more like the definition of absolute value, which is based on the fact that $\sqrt{x}$ will always return a positive value. – GeoffDS Oct 30 '12 at 18:08
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    @Graphth The definition of $\vert \cdot \vert$, I use is $$\vert x \vert = \begin{cases} x & \text{if } x\geq0 \ -x & \text{if } x < 0\end{cases}$$ –  Oct 30 '12 at 18:10
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    In reality, we know what a square root is before we define $\sqrt{}$, and so we define that symbol to mean the positive square root. But, I've never seen any one define what $\sqrt{}$ means by $\sqrt{x^2} = |x|$. I guess it's just a more complicated way of saying the same thing. Since your reputation is 7 times mine, I will defer to you :) – GeoffDS Oct 30 '12 at 18:59
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You are correct that $(-2)^2 = 4$.

The idea is that we want $\sqrt{x}$ to be a continuous, single valued function. But as you noted there are two possibile values of $\sqrt{x}$ for each $x$, so we have to choose a particular branch and that is what we call $\sqrt{x}$.

So the standard choice is to just take positive $\sqrt{x}$ for every $x$.

Deven Ware
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