# Square root of simple binomial function [duplicate]

Let's say I have the following formula:

$$\sqrt{a^2-2ab+b^2}=\sqrt{(a-b)^2}=\sqrt{(b-a)^2}$$

When do I know which one of the following I should use?:

$$\sqrt{(a-b)^2}=a-b\qquad\text{ or }\qquad \sqrt{(b-a)^2}=b-a$$

## marked as duplicate by Namaste algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 7 '18 at 23:55

• Square roots are non-negative by definition, so you use whichever one is non-negative. Equivalently, you use $|a - b|$. – Qiaochu Yuan Jun 12 '11 at 0:26
• – Eric Naslund Jun 12 '11 at 0:50
• @Qiaochu: Shouldn't this be an answer? Seeing as it is the correct answer... – trutheality Jun 12 '11 at 0:53
• I think "quadratic function" may be better than "binomial function". – Jack Dec 15 '11 at 6:53

Let $x=a-b$, then $-x=b-a$. Then $$\sqrt{(b-a)^2}=\sqrt{(-x)^2}=\sqrt{x^2}=\sqrt{(a-b)^2}$$ Now matter what real number you take for $a$ and $b$, you always have $$\sqrt{(b-a)^2}=\sqrt{(a-b)^2}$$
So in my opinion, your question may be somewhat misleading for yourself. You are actually asking when $$\sqrt{x^2}=x$$ and when $$\sqrt{x^2}=-x.$$ So what you need is the definition of square root: for all real numbers $x$, $$\sqrt{x^2}=|x|$$ Now you can go on the argument yourself.
$\sqrt{(a-b)^2} = |a-b| = |b-a| = \sqrt{(b-a)^2}$.