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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!

Git Gud
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Ina
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    $\sqrt{x^2} = |x|$ 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. – Guy Mar 15 '14 at 11:40
  • If it's a definition, why square root of 9 is +/- 3 and not +3? – Ina Mar 15 '14 at 11:41
  • It is +3, not $\pm 3$ – Guy Mar 15 '14 at 11:42
  • 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? – Ina Mar 15 '14 at 11:42
  • @Ina provide an example. – Guy Mar 15 '14 at 11:43
  • F.e. here http://www.purplemath.com/modules/radicals.htm, section Simplifying Square-Root Terms, the case with negative values is ignored. Why? – Ina Mar 15 '14 at 11:47
  • @GitGud okay. deleting. – Guy Mar 15 '14 at 11:57
  • By definition , $;\sqrt{x^2}=|x|;$ and this, as already commented, answer all your doubts in this question. Now, if we can't explicitly know what square root we want, or if both (positive-negative) are equally valid, we write for example $;\pm\sqrt{9}=\pm3;$ to express both possibilites, but still $;\sqrt 9=3;$...again, by definition . – DonAntonio Mar 15 '14 at 12:05

4 Answers4

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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.

MPW
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  • 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? – Ina Mar 15 '14 at 12:17
  • @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). – hmakholm left over Monica Mar 15 '14 at 12:24
  • 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? – Ina Mar 15 '14 at 12:29
  • 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. – MPW Mar 15 '14 at 13:42
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    @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. – hmakholm left over Monica Mar 15 '14 at 13:50
  • ... The parabola comes in only when you imagine doing the such-and-such arithmetic on numbers that are NOT the one I'm thinking of. Then you can plot the results and see that they form a parabola (and doing so may indeed be helpful for figuring out which number I'm thinking of) but when you come back and tell me what you think my number is, you're not speaking about any parabolas -- only about what the number I'm thinking of is. – hmakholm left over Monica Mar 15 '14 at 13:52
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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.

hmakholm left over Monica
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  • 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 – Ina Mar 15 '14 at 12:12
  • @Ina: Yes, exactly. – hmakholm left over Monica Mar 15 '14 at 12:17
  • 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 "$x=(-b\pm\sqrt{b^2-4ac})/2a$". – MPW Mar 15 '14 at 13:26
  • @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. – hmakholm left over Monica Mar 15 '14 at 13:36
  • @HenningMakholm: I accept that. My comment was really intended for Ina's benefit, not a correction of your explanation. Apologies. – MPW Mar 15 '14 at 13:50
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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

PossibilityZero
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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 $3$.

Klaas van Aarsen
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