# Expanding Square Roots, Why No Negative?

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 Tamaroff Jun 9 '14 at 0:58
• To further elaborate on Pedro's comment, you might find this question and its answers helpful. – Michael Albanese Jun 9 '14 at 1:01
• 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 9 '14 at 1:02
• Thanks, fixed that. I understand now. Thank you all! – Z. Bornheimer Jun 9 '14 at 17:05

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