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This question already has an answer here:

The answer to my question might be obvious to you, but I have difficulty with it.

Which equations are correct:

$\sqrt{9} = 3$

$\sqrt{9} = \pm3$

$\sqrt{x^2} = |x|$

$\sqrt{x^2} = \pm x$

I'm confused. When it's right to take an absolute value? When do we have only one value and why? When two and why?

Thank you very much in advance for your help!

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marked as duplicate by 6005, Daniel W. Farlow, naslundx, MickG, Henrik Sep 9 '16 at 20:07

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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    $\begingroup$ $\sqrt{x^2} = |x|$ It is simply defined to be so. $\sqrt{x}$ is the positive solution of $x=y^2$ I mean the second statement is a definition. The first statement simply follows. $\endgroup$ – Guy Mar 15 '14 at 11:40
  • $\begingroup$ If it's a definition, why square root of 9 is +/- 3 and not +3? $\endgroup$ – Ina Mar 15 '14 at 11:41
  • $\begingroup$ It is +3, not $\pm 3$ $\endgroup$ – Guy Mar 15 '14 at 11:42
  • $\begingroup$ because 9 can be made of -3 or 3, right? But sometimes in calculation people just use square root of 9 is 3. What if it's -3? $\endgroup$ – Ina Mar 15 '14 at 11:42
  • $\begingroup$ @Ina provide an example. $\endgroup$ – Guy Mar 15 '14 at 11:43
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The confusion about the sign is understandable. The square root symbol applied to a positive number always yields a positive number (disregarding the case of zero for the sake of simplicity here). The problem arises when you don't know ahead of time whether $x$ is positive or not. It is true that one of the numbers $x$ and $-x$ must be positive, though. So you can write with certainty that $$\sqrt {x^2}=|x|$$ since $|x|$ is precisely the one of these two numbers that is positive--it's just another way to say the same thing more concisely.

It is also true that "either $\sqrt{x^2}=x$ or $\sqrt{x^2}=-x$" is true, which is often abbreviated as "$\sqrt{x^2}=\pm x$". But be very careful what this says. It is a disjunction, a compound statement that at least one of the two component statements must be true. It does not say that both must be true. So it is also correct to write $$\sqrt{x^2}=\pm x$$ if you understand that it means "or" but not necessarily "and".

So to answer your question: they are all correct.

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  • $\begingroup$ When we look for zeros of a quadratic equation, on the way to quadratic formula, we have this: $(x+\frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} => x+\frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$ => $x = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} - \frac{b}{2a}$. We "accept" those two zeros, don't we? We know the graph crosses x-axis in two points, if both zeros exist. Why is it "and" not "or" in this case? $\endgroup$ – Ina Mar 15 '14 at 12:17
  • $\begingroup$ @Ina: It is "or" in that case too. If you know about $x$ that it satisfies the quadratic equation, what you can deduce that $x$ is one root OR $x$ is the other root. You can't deduce that $x$ equals both roots simultaneously. (And conversely, if $x$ is one root OR $x$ is the other root, then you know it satisfies the equation). $\endgroup$ – Henning Makholm Mar 15 '14 at 12:24
  • $\begingroup$ it means, i cannot say that the parabola crosses the x-axis in the two points? so there must be another 2 points that correspond to the found ones, which would tell us where a parabola probably crosses the x-axis. Right? So, probably it's situated to the right of the y-axis, or to the left of the y-axis, depending on which of the zeros are true. So.. does it mean, a given quadratic equation with two possible zeros is a equation for two different parabolas? $\endgroup$ – Ina Mar 15 '14 at 12:29
  • $\begingroup$ The point is that you can only plug one value of $x$ into the equation at a time. So $x$ satisfies $x^2-9=0$ precisely if either $x=3$ or $x=-3$, which means that the parabola meets the $x$-axis at the two points $(3,0)$ and $(-3,0)$. I think you are having a problem with letting $x$ assume different values in different circumstances. It isn't equal to both values at the same time. The equation is just a condition that may or may not be satisfied by a given number; the numbers that satisfy it are the $x$-intercepts. $\endgroup$ – MPW Mar 15 '14 at 13:42
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    $\begingroup$ @Ina: No that is confused. Neither the question nor the answer is (explicitly) about parabolas. Think of the quadratic equation as a riddle: "I'm thinking of a number, and when you do such-and-such arithmetic on it you get zero. What is the number?". Then you go away and do some math, and you come back and say "Your number is 5 OR your number is 17". You don't say "Your number is 5 AND your number is 17", that would be impossible. Sure, you can also say "The only possible values of your number are 5 AND 17", but that is not a logical "and", merely a list-forming one. $\endgroup$ – Henning Makholm Mar 15 '14 at 13:50
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By definition the square root of a number is the positive number whose square is the original number. So we have $\sqrt9=3$ and $\sqrt{x^2}=|x|$ and no doubt about either.

There is no number whose square root is $-3$ (even if we move to complex numbers and consider principal square roots).

What can create confusion is that we sometimes have an equation such as $$ x^2 = 9 $$ and say something like "now let's take the square root on both sides" to get $$ x = \pm 3 $$ which can look like we're saying taking the square root of $9$ gives $\pm 3$. But what really happens is that the square roots give us $$ |x| = 3 $$ and then there's an implicit invisible step that replaces the absolute value sign with a $\pm$ to get $x=\pm 3$ instead.

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  • $\begingroup$ so, it's like: $x^2 = 9$ <=> $\sqrt{x^2} = \sqrt{9}$ <=> $\sqrt{x^2} = |x| = \sqrt{9} = 3$ => |x| = 3 => x = 3 or x = -3 $\endgroup$ – Ina Mar 15 '14 at 12:12
  • $\begingroup$ @Ina: Yes, exactly. $\endgroup$ – Henning Makholm Mar 15 '14 at 12:17
  • $\begingroup$ This is all shorthand for what's really going on: "$x^2=9$" is equivalent to "$(x+3)(x-3)=0$", which in turn is equivalent to "$x+3=0$ or $x-3=0$" and so to "$x=3$ or $x=-3$", i.e., "$x=\pm 3$". Exactly the same line of reasoning gets you from "$ax^2+bx+c=0$" to "$x=(-b\pm\sqrt{b^2-4ac})/2a$". $\endgroup$ – MPW Mar 15 '14 at 13:26
  • $\begingroup$ @MPW: No, I don't think that is "what's really going on". The factoring approach you show is a different way of reaching the same conclusion. $\endgroup$ – Henning Makholm Mar 15 '14 at 13:36
  • $\begingroup$ @HenningMakholm: I accept that. My comment was really intended for Ina's benefit, not a correction of your explanation. Apologies. $\endgroup$ – MPW Mar 15 '14 at 13:50
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In the real numbers, $\sqrt x$ is defined to be positive.

In the complex numbers, $\sqrt z$ is a multivalued function that indeed yields 2 values. In that case we have a principal value of $\sqrt 9$ that is $3$.

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  • $\begingroup$ $\sqrt{x}$ is defined to be non-negative. $\endgroup$ – Michael Hoppe Mar 15 '14 at 13:19
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The mathematical symbol √ refers to positive number of the two possible square roots.

If the question is written as "What is the square root of 9?", then the answer is both 3 and -3.

However, if the question is "Evaluate √9," the answer would only be 3. Consequentially, -√9 = -3

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