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This question already has an answer here:

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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marked as duplicate by Namaste algebra-precalculus Feb 7 '18 at 23:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Square roots are non-negative by definition, so you use whichever one is non-negative. Equivalently, you use $|a - b|$. $\endgroup$ – Qiaochu Yuan Jun 12 '11 at 0:26
  • $\begingroup$ Related: math.stackexchange.com/questions/41878/… $\endgroup$ – Eric Naslund Jun 12 '11 at 0:50
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    $\begingroup$ @Qiaochu: Shouldn't this be an answer? Seeing as it is the correct answer... $\endgroup$ – trutheality Jun 12 '11 at 0:53
  • $\begingroup$ I think "quadratic function" may be better than "binomial function". $\endgroup$ – Jack Dec 15 '11 at 6:53
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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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