Expanding Square Roots, Why No Negative?

Question

I haven't thought through algebra in a while and the last explanation I received of this seemed arbitrary. I hope I can get some clarification here.

I understand that $\sqrt{+a} = \pm b$. Here's something I don't however.

Evaluate using FOIL: $(\sqrt{8} - \sqrt{2})(\sqrt{8} - \sqrt{2})$

Evaluation yields: $8 - \sqrt{16} - \sqrt{16} + 2 == 8 - 2\sqrt{16} +2$

In this case, why must $-2\sqrt{16} = 8$ and not $-8$?

Thanks!

We always agree that $\sqrt a$ denotes the unique positive real number such that $x^2=a$. Hence $\sqrt{16}=4$. — Pedro, Jun 09 '14 at 00:58
To further elaborate on Pedro's comment, you might find this question and its answers helpful. — Michael Albanese, Jun 09 '14 at 01:01
Your FOIL got the wrong sign on the last term, it confused me at first how the multiplication of two positive numbers could be negative... — abiessu, Jun 09 '14 at 01:02

score 2 · Answer 1 · edited Jun 12 '20 at 10:38

2

Question is answered in the comments under my original post:

We always agree that $\sqrt a$ denotes the unique positive real number such that $x^2=a$. Hence $\sqrt{16}=4$. -- Pedro Tamaroff

To further elaborate on Pedro's comment, you might find this question and its answers helpful. -- Michael Albanese

edited Jun 12 '20 at 10:38

Community

1

answered Jun 09 '14 at 17:06

Z. Bornheimer

51

Expanding Square Roots, Why No Negative?

1 Answers1