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We know that the square root function ($\sqrt{}$) always produces non-negative numbers. Why then, does the square root function produce a negative value here?

$$x^2=4$$

$$\implies \pm \sqrt{x^2}=2$$

$\implies \sqrt{x^2}=2...(i)$, or, $\implies \sqrt{x^2}=-2...(ii)$

In equation (ii), the square root function produces a negative value. I can't even graph equation (ii) in desmos. Why is this happening?

tryingtobeastoic
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  • The square-root function always producing non-negative results is a convention, and conventions can be followed or not followed depending on convenience. – user3716267 Sep 03 '21 at 15:40
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    You can alternatively approach it as follows:

    $$x^{2} = 4 \Longleftrightarrow x^{2} - 4 = 0 \Longleftrightarrow (x-2)(x+2) = 0$$

     – user0102 Sep 03 '21 at 15:41
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    You shouldn't have any “$\pm$” in the first step to begin with! The equation $x^2=4$ is equivalent to $\sqrt{x^2} = 2$, or in other words $|x|=2$. You could include case (ii), but it's always false, so “(i) or (ii)” is in fact the same thing as just “(i)”. – Hans Lundmark Sep 03 '21 at 19:35

1 Answers1

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What does the equation $x^2 = 4$ mean? On its face, it states that we have some $x$ (we'll assume it's a real number), such that $x * x = 4$.

There are two such real numbers. Clearly, $2 * 2 = 4$. But also, $(-2) * (-2) = 4$. So, the equation $x^2 = 4$ has two solutions.

This is true regardless of how we define the square-root function. Since we tend to require that functions map each input to only one output, the canonical "square-root function" takes only non-negative values. But this is merely a convention; one consequence of this convention is that we must use the fairly cumbersome $\pm$ symbol to indicate the presence of both solutions.

user3716267
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