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I had the following question:

Simplify $y = \sqrt{a^4b^2}$

It did not say anything about the domain, so I assumed $x \in (-\infty, +\infty)$

It seems correct to say that the answer is: $y = \pm a^2b$

However, the answer is apparently $y = a^2b$. Why is this?

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4  
That answer is not right in my view, there should be an absolute value on that $b$ –  imranfat Apr 2 at 15:46
    
Wouldn't that only be true if it were the principal square root? –  Jason Apr 2 at 15:47
3  
The square-root operation itself is defined to produce only the positive square root. When we solve equations such as $ \ x^2 = 5 \ , $ this has two admissible solutions, so we must write the "plus-or-minus" symbol in front of the square-root operation to indicate that, since the operation only gives the positive result: $ \ x = \pm \sqrt{5} \ . $ –  RecklessReckoner Apr 2 at 15:47
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$a^2|b|$ is the right answer. –  kmitov Apr 2 at 15:51
    
@imranfat I'll agree that there is more than a little sloppiness in the way such expressions are discussed. (A recent thread on fractional exponents sure brought that out; I learned something from that myself.) I suspect in the context of the course that the intended domain is positive real numbers, but for all reals, I agree $ \ a^2 \ |b| \ $ would be correct; here, $ \ \pm a^2 b \ $ would be unclear as to when to use which sign. –  RecklessReckoner Apr 2 at 15:55

2 Answers 2

up vote 4 down vote accepted

The true answer is that $\sqrt{a^4b^2}=|a^2b|=a^2|b|$, since the notation $\sqrt{\phantom{bb}}$ is defined to mean the positive square root of the argument.

If someone claims that the answer is $a^2b$, it must be because they think (wrongly or rightly, depending on the context) that $b$ is nonnegative.

Your $y=\pm a^2b$ is shorthand for $y=a^2b \lor y=-a^2b$ -- and this is a "simplification" of the original equation in the sense that $$ y=\sqrt{a^4b^2} \implies (y=a^2b \lor y=-a^2b) $$ but the implication in the other direction doesn't hold, so by writing $y=\pm a^2b$ you have increased the set of $(y,a,b)$ triples that satisfy the condition, which is probably not what you intended to do.

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Jason. The square root of 9 is by definition 3 and not -3. However, the equation $x^2=9$ has two solutions, being 3 and -3. I hope that answers your question as of why the squareroot of some term is not +/- Then, the squareroot of $x^2$ is $|x|$ because it is indeed $x$ if $x>0$ and it is $-x$ if $x<0$ (zero can be put in either choice) You labeled your question as a precalc question so if you have a precalc book, this should be somewhere in there. It is one way to introduce the absolute value function

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