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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?:

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

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    $\begingroup$ Square roots are non-negative by definition, so you use whichever one is non-negative. Equivalently, you use $|a - b|$. $\endgroup$ Jun 12, 2011 at 0:26
  • $\begingroup$ Related: math.stackexchange.com/questions/41878/… $\endgroup$ Jun 12, 2011 at 0:50
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    $\begingroup$ @Qiaochu: Shouldn't this be an answer? Seeing as it is the correct answer... $\endgroup$ Jun 12, 2011 at 0:53
  • $\begingroup$ I think "quadratic function" may be better than "binomial function". $\endgroup$
    – user9464
    Dec 15, 2011 at 6:53

2 Answers 2

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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.

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$\sqrt{(a-b)^2} = |a-b| = |b-a| = \sqrt{(b-a)^2}$.

Without the absolute value sign, the identity is correct only when the larger of the two numbers comes first in the subtraction, since the radical refers to the nonnegative square root.

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