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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!

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    $\begingroup$ We always agree that $\sqrt a$ denotes the unique positive real number such that $x^2=a$. Hence $\sqrt{16}=4$. $\endgroup$
    – Pedro ♦
    Jun 9, 2014 at 0:58
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    $\begingroup$ To further elaborate on Pedro's comment, you might find this question and its answers helpful. $\endgroup$
    – Michael Albanese
    Jun 9, 2014 at 1:01
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    $\begingroup$ 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... $\endgroup$
    – abiessu
    Jun 9, 2014 at 1:02
  • $\begingroup$ Thanks, fixed that. I understand now. Thank you all! $\endgroup$
    – Z. Bornheimer
    Jun 9, 2014 at 17:05

1 Answer 1

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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

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