## Online Course Discussion Forum

### Why can't there be a negative root for a radical function?

This is a very good question. We need to be very clear about the meanings of the definition of square roots and the notation we use for them.

Each positive number $a$ has two square roots, one positive and one negative. The number $0$ has only one square root, of course, which is $0$. When we write the notation $\sqrt{a}$, it represents the nonegative one. So the square roots of positive number $a$ are $\sqrt{a}$ and $-\sqrt{a}$. This way, there is no confusion for the expression.

Therefore, whenever you see the expression such as $\sqrt{x^2-3x+2}$, it means the nonnegative square root of $x^2-3x+2$. It is the same with radicals with even index, such as $\sqrt[4]{m+1}$ and $\sqrt[20]{y^3-3}$: it is required that the value underneath the radical is nonnegative, and the result is also nonnegative. So $\sqrt[4]{16}=2$, and $\sqrt[6]{-64}$ is undefined in real numbers. For radicals with odd index, there is no such restriction: $\sqrt[3]{-8}=-2$.

Thanks a lot for the clear explanation! Sorry for pestering with one following question, referring to this line of reply: So $\sqrt[4]{16}=2$164=2, and $\sqrt[6]{-64}$, though, undefined in real numbers? 2 to the fourth power is 16, and so does -2, which looks mathematically plausible. I do understand that there is no solution for $\sqrt[6]{-64}$ as its index is not an odd number, but so does the former one.

Again, thanks a lot for the quick reply! Areteem courses are gearing me up for the upcoming AMC 10 Exam. Thanks to Instructor John and David for the fantastic tuition

(Edited by Areteem Professor - original submission Monday, August 23, 2021, 5:51 AM)

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