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$\sqrt{x}$$\sqrt{x}$ = $x$ but $\sqrt{x^2}$ = $|x|$. Why is this?

I'm just learning algebra again after many years and I can't seem to figure out why this is. I'm sure this is trivial but if someone could explain it it would help me a lot. Thanks!

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The domain of your first expression is $x>=0$ For the second the domain is all reals, so look at the case of$x$ positive or negative separately, match with absolute value result. –  coffeemath Oct 17 '12 at 18:31

2 Answers 2

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By definition $\sqrt x$ is the unique non-negative real number $y$ such that $y^2=x$, so $\sqrt x\cdot\sqrt x=x$ is true by definition.

Now apply the definition to $\sqrt{x^2}$: $\sqrt{x^2}$ is the unique non-negative real number $y$ such that $y^2=x^2$. If $x=0$, the only real number whose square is $x^2$ is $0$, so of course $\sqrt{x^2}=0=|0|$. If $x\ne 0$, there are always two real numbers $y$ such that $y^2=x^2$: one of them is $x$, and the other is $-x$. Exactly one of these two is positive. Since we don’t know whether $x$ is positive or not, we don’t know which of them is positive, but we know that whichever it is, it’s $|x|$. Therefore $|x|$ is the unique positive real number such that $|x|^2=x^2$, and by definition $\sqrt{x^2}=|x|$.

As an example, suppose that $x=-3$. Then $x^2=9$, and the two real numbers whose squares are $9$ are $-3$ (i.e., $x$) and $3$ (i.e., $-x$). The non-negative one is $3=-(-3)=|-3|$. Had we started with $x=3$, $x^2$ would still have been $9$, and we’d still have wanted the positive one of $x$ and $-x$, but this time that would be $x$, not $-x$. It’s still true, however, that $|x|=|3|=3$, the one that we want.

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Thanks, that explained it :) –  jon Oct 17 '12 at 18:46
    
@jon: You’re welcome. Good luck with the algebra! –  Brian M. Scott Oct 17 '12 at 18:48

I think your source of confusion here comes from thinking of $\sqrt{x}$ as a number rather than a function. Often times, we write something as $\sqrt{x}$ to denote the squareroot(s) of the number $x$. However, in many other situations, we also write $\sqrt{x}$ to talk about the function which sends the number $x$ to its square root.

In particular, in your example, the first instance $\sqrt{x}\sqrt{x} = x$, is most likely talking about the number $\sqrt{x}$. Indeed, by definition of the number $\sqrt{x}$, we have that equation.

In the second instance, $\sqrt{x^2} = |x|$, it is more appropriate to think about $\sqrt{x}$ as a function, which is composed with the function $x^2$. In that case, $\sqrt{x}$ must be strictly positive, since there are no real numbers whose square roots are negative!

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You don't even talk about the sign of $x$ in the two examples of the OP. –  enzotib Oct 17 '12 at 18:40
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I agree with @enzotib: this doesn’t address the OP’s question. –  Brian M. Scott Oct 17 '12 at 18:41
    
First off, $\sqrt{x}$ is a number, once a particular $x$ is chosen. But I agree with Cheng that part of the definition of $\sqrt{x}$ is that it's square be x. But Chengs third paragraph surprized me: You can't say that the fact that $\sqrt{x}$ must be strictly positive (or better:nonnegative) has anything to do with the calculation of $\sqrt{x^2}$. Indeed the latter is to compose the two functions such that we first find $x^2$, and then throw that result into sqrt(x) as an input to it. –  coffeemath Oct 17 '12 at 19:17

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