# Is √64\sqrt{64} considered 88? or is it 8,−88,-8?

Last year in Pre-Algebra we learned about square roots. I was taught then that
$$√64=8\sqrt{64}=8$$ and $$√100=10\sqrt{100}=10$$, which I understood and accepted. I was also taught that $$±√64=8,−8\pm\sqrt{64} = 8,-8$$ because both of those numbers squared is 64, which I also get.
But this year, with a new school and teacher in a different state, our teacher is telling us that:
$$√64=8,−8\sqrt{64}=8,-8$$ and $$±√64\pm\sqrt{64}$$ also is $$8,−88,-8$$. The way to get the positive root of something is:
$$+√64=8+\sqrt{64}=8$$

And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $$nn$$ is defined as
$$y2=ny^2=n$$ I see where he is coming from.

Upon researching this Wikipedia says:

For example, $$44$$ and $$−4−4$$ are square roots of $$1616$$ because $$42=(−4)2=164^{2} = (−4)^{2} = 16$$

And Wolfram MathWorld says:

Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $$99$$ are $$−3-3$$ and $$+3+3$$

But on the other side, Wolfram Alpha, when given “The square root of 9” gives only 3.

So, which is right? Is $$√64\sqrt{64}$$ considered $$88$$? or is it $$8,−88,-8$$?

Your new teacher is wrong. $\sqrt{\cdot}$ is the principal square root operator. That means it returns only the principal root — the positive one. $\sqrt{64}=8$. It does NOT equal $-8$.
On the other hand, the equation $64=x^2$ DOES have $2$ solutions: $x=8$ or $x=-8$. Thus both $8$ and $-8$ are square roots of $64$.