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This question already has an answer here:

Let's say I am solving an equation, and end up with this:

x^2 = 16

The solutions will be

x=4 or x=-4

That makes sense. But when I have this:

x = √16

The only solution is

x=4

Because roots do not have negative outcomes. But when I solve x^2 = 16, what I actually do is take the root of 16 and that is my answer. Can anyone explain this? And why do square roots only have positive answers?

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marked as duplicate by David K, Did, Aloizio Macedo, Harish Chandra Rajpoot, Marconius Oct 15 '15 at 0:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The symbol refers to one of the square roots. The fact that we call it "the" square root" instead of "the positive square root when one exists" is inaccurate laziness but coloquially understood.

So the solutionS to $x^2 = 16$ are both 4 (which is $\sqrt {16}$) AND -4 (which is $-\sqrt{16}$) and we frequently write it as $\pm\sqrt{16}$)

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Note: $x^2 = 16$ and $x = \sqrt{16}$ are two different statements altogether. They are related statements but still different. In getting from $x^2 = 16$ to $x = \sqrt{16}$, we could just as easily have gone to the statement $-x = \sqrt{16}$.

If we want to be pedantic about it, the proper statement should have been $\pm x = \pm \sqrt{16}$ however these four statements are redundant, so simply $x = \pm \sqrt{16}$ suffices. (Although techinally $\pm x = \sqrt{16}$ is equally valid.)

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It's a convention to make $\sqrt{} $ a function. A function can only have one value for each input, so for every positive real $x$ we define $\sqrt{x}$ as the positive real number such that $(\sqrt{x})^2=x$.

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Remember this always : If $y^2=x$ then, $\sqrt{x}=|y|$,. Since y satisfies both positive and negative values we have them.

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