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I have a question regarding this video

https://www.khanacademy.org/math/algebra/exponent-equations/exponent-properties-algebra/v/adding-and-simplifying-radicals

He said that $\sqrt {x^2}=|x|$ has to be the absolute value of $x$ because $x$ can be positive or negative, which makes sense but why is $\sqrt 4= 2$? Shoudln't it be $|2|$ too? Because $-2 \cdot -2 = 4$ and $2 \cdot 2 = 4$.

Later
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Maik Klein
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I'd make this a comment if I had the reputation, but as it stands, just think about what $|2|$ is.

Branimir Ćaćić
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$\left | 2 \right |=2$. Indeed $\sqrt{x^2}=\left | x \right |$ since $\sqrt{x^2}=x$ only for non-negative values of $x$. Consider for example this case:

$\sqrt{9}=\sqrt{(-3)^2}\neq -3$. But $\left | -3 \right |=3$ which is the correct answer.

EricAm
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  • $\sqrt{9}=\sqrt{(-3)^2}\neq -3$ I am still a little bit confused, Wouldn't -3 and +3 both be correct results? I explained it to me the following way in this example $\sqrt{x^5}$ Because it is $x^5$ it has to be positive, because the square root of a negative number is undefined. But if we have $\sqrt{x^4}$ then x could be negative or positive. But maybe I am just a little bit confused. – Maik Klein Jan 01 '13 at 17:40
  • $(-3)^2=9$ since$-3\times -3=9$. – Basil R Jan 01 '13 at 17:54
  • @MaikKlein: What is your problem. Tell me explicitly. :-) – Mikasa Jan 01 '13 at 18:02
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    @Maik, when we write $\sqrt9$, we don't mean "any one of the two numbers whose square is $9$", we mean "the positive number whose square is $9$". That is the universally agreed-upon meaning of the $\sqrt{\cdot}$ symbol. See this previous question: Square roots — positive and negative. –  Jan 01 '13 at 18:07
  • Ah okay now it makes sense! That's the reason why we say $\sqrt{x^2} = | x$ | because the output of the function squareroot can only have 1 output , otherwise it wouldn't be a function. And the output is just defined to be positive, that why we have to force the x to be positive. Okay thank you very much – Maik Klein Jan 01 '13 at 18:52
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Remember the definition of $|x|$: $$ |x| = \left\{ \begin{array}{ll} x & \quad x \geq 0 \\ -x & \quad x < 0 \end{array} \right. $$ So what is $|-2|$ according to this definition? It is $|-2|=-(-2)=2$

Mikasa
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Basically, the square root function is the inverse of the square function. However, we know that only bijections are invertible and $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^2$ is neither one-one nor onto, and thus, not a bijection.

Note that it being not a bijection, doesn't/can't have an inverse.

However, if we restrict the domain and the codomain of this function to the set of non-negative real numbers, $\mathbb{R_+}$, then $f:\mathbb{R_+}\to\mathbb{R_+}$ defined by $f(x)=x^2$ is a bijection and invertible. Its inverse is the famous square root function $f^{-1}:\mathbb{R_+}\to\mathbb{R_+}$.

Consequently, $\sqrt{x^2}=x$ only if $x\geq0.$

F. A. Mala
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