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For example, $\sqrt{4}$. I've asked a bunch of people and I get mixed answers all the time, as to whether it is $-2$ and $+2$ or just $+2$.

How about if there's a negative in front of the square root sign, for example, $-\sqrt{4}$? Would that still be plus or minus or just minus?

pjs36
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John
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    Square root is always positive. It is defined to be positive. – Apurv May 10 '16 at 02:20
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    So the square root of 4 is just positive two? – John May 10 '16 at 02:22
  • I have an addition question. Say that you have sqrt((-5)^2). Shouldn't the answer be -5 because the exponent inside the square root cancels with the square root itself and just leaves the -5? Why do I get just 5 when I plug it in my calculator. – John May 10 '16 at 02:23
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    Note that $\sqrt {x^2}=|x|$, which is always positive. You cannot simply cancel the exponent. You have to make sure that you get a positive number. The graph of the square root function ($y=\sqrt x$) is defined to stay in the first quadrant. – Apurv May 10 '16 at 02:24
  • comment edited. – Apurv May 10 '16 at 02:26
  • What's the procedure you would use to solve sqrt((-5)^2)? – John May 10 '16 at 02:39
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    $\sqrt {(-5)^2}=|-5|=5$, as per the definition I just gave in the previous comment. – Apurv May 10 '16 at 02:43
  • Are you automatically distributing the squared to the (-5)? – John May 10 '16 at 02:47
  • No one is distributing anything. Just follow the definition. $\sqrt {x^2}=|x|$. – Apurv May 10 '16 at 02:53
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    First Question: It is a matter of mathematical convention that $\sqrt{x}\ge0$ for all $x\ge0$, so that $\sqrt{x}$ is well-defined and has a single value.

    Second Question: $(-5)^2=25$ hence $\sqrt{(-5)^2}=\sqrt{25}=5$.$.

     – Mirko May 10 '16 at 02:53
  • @Mirko Yes, so you simplified the (-5)^2 first, then you square rooted, correct? I have not gotten to absolute values with square roots yet. – John May 10 '16 at 02:56
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    You just got to exactly this. If $x\ge0$ then $\sqrt{x^2}=x=|x|$, just like $\sqrt{5^2}=5=|5|$. If $x<0$ then $-x>0$ hence $\sqrt{x^2}=\sqrt{(-x)^2}=-x=|x|$, just like $\sqrt{(-5)^2}=\sqrt{25}=5=-(-5)=|-5|$. – Mirko May 10 '16 at 03:01

1 Answers1

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This is a common source of confusion, because people don't clearly separate in their minds a few related but different situations.

  • Does $\sqrt4$ mean $\pm2$ or just $2$? The answer: it means $2$. By definition, the $\sqrt{\cdot}$ function always evaluates to a nonnegative number (as long as it's being applied to a nonnegative number; otherwise it's not defined at all). The definition of $\sqrt x$ is: the nonnegative number $y$ such that $y^2=x$.
  • Does $(\sqrt x)^2$ always equal $x$? (We'd better assume that $x$ itself is nonnegative for the notation to make sense.) The answer: yes. This is exactly the definition of the $\sqrt\cdot$ function, as described above.
  • Does $\sqrt{x^2}$ always equal $x$? The answer: no, since "the nonnegative number whose square is $x^2$" is not always $x$. But $|x|$ is a nonnegative number and its square is $x^2$; therefore $\sqrt{x^2}=|x|$.
  • Suppose that $x^2=4$; does that mean that $x=\pm2$ or just $x=2$? The answer: it means $x=\pm2$. If we apply the square root function to both sides of the equation $x^2=4$, we get $|x|=2$, which is equivalent (by the definition of absolute value) to $x=\pm2$.
Greg Martin
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