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If $x=9$, then can we write $\sqrt{x}=\pm3$ or only $\sqrt{x}=3$.

I am confused as square root always gives positive number.

But the irony is that if we have $a^2=9$, then we write $a=\pm3 \text { where $a=\sqrt{a^2}$ }$

user3290550
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  • I'm confident this has been addressed before, so I'll find a proposed duplicate. But the gist of the matter is that while $-3$ is "a square root" of $9$, the notation $\sqrt 9$ means the principal square root of $9$, namely $3$. It is convenient to have this sort of meaning so that $\sqrt x$ has a functional dependence on $x$, at least for the simple case $x\ge 0$. – hardmath Nov 04 '19 at 02:09
  • The square root function has a range of positive values only. – Sina Babaei Zadeh Nov 04 '19 at 02:12
  • Regarding your "irony", if we have $a^2=9$ then we write the two solutions $a=\pm \sqrt 9 = \pm 3$. – hardmath Nov 04 '19 at 02:16
  • @hardmath, but $a=\sqrt{a^2}$, how can $a$ be negative – user3290550 Nov 04 '19 at 02:18
  • Given $a^2=9$, there are two solutions to the quadratic equation. The notation $\sqrt 9$ meaning the positive square root of $9$ makes the notation specific, and allows us to identify both distinct roots, $+\sqrt 9$ and $-\sqrt 9$. – hardmath Nov 04 '19 at 02:21
  • ok, for $x=9$, should we write $\sqrt{x}=\pm3$ or $\sqrt{x}=3$ – user3290550 Nov 04 '19 at 02:23
  • If $x=9$, then $\sqrt x = 3$. Right? Given a value of $x\ge 0$, there exists a unique positive square root of $x$, and that is what we mean by $\sqrt x$. Also, the meaning of $\sqrt{a^2}$ is the same as $|a|$, if $a$ is a real number. – hardmath Nov 04 '19 at 02:26
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    @user3290550: The equality $a=\sqrt{a^2}$ is only true if $a \ge 0$. In general, $\sqrt{a^2}=|a|$. – Hans Lundmark Nov 04 '19 at 07:08

4 Answers4

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Note that $x^2=9$ has two solutions namely $x=\sqrt 9=3$ and $x=-\sqrt 9 = -3$

When we write $x=\pm 3$ we mean that both $3$ and $-3$ are solutions and it does not mean that they are equal.

Thus we have $x=\pm \sqrt 9$ which means there are two solutions and they are opposite to each other.

In general for a real number $x$ we have $$\sqrt {x^2}= |x|$$ which results in $$-\sqrt {x^2}= -|x|$$

Mohammad Riazi-Kermani
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The square root, as a function, is defined to give only the positive value, because functions are allowed to have a unique image for every element in the domain.

However, if you think of the square root as a relation (functions are relations with special constraints set on them), then you can relate an element to both "square roots"

Francisco José Letterio
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  • see I am asking a very simple question if $a^2=9$, then we write $a=\pm3$, but $a=\sqrt{a^2}$, how can $a$ be negative if square root of a number is always positive. – user3290550 Nov 04 '19 at 02:20
  • it depends of what you want the square root to be, or what you call the square root of a number. As a function/map $sqrt:\mathbb R \rightarrow \mathbb R $ then you can only take a positive value, because of the definition of a function. However, that's not the same as asking for the solutions of $a^2 = 9$, because those solutions are based on the notion that the square root is a function and thus the mapping of an element is unique – Francisco José Letterio Nov 04 '19 at 02:30
  • in other words, the function $sqrt(x)$ is always positive, which is not the same as saying that THE square root (there is no "THE square root" because, except for 0, it's not unique) is always positive. There is a distinction between "the function that maps a positive integer to its positive square root" and "the relation that relates a positive integer with all of its square roots" – Francisco José Letterio Nov 04 '19 at 02:34
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    @user3290550: It is false that “$a=\sqrt{a^2}$” always holds. What we have is $\sqrt{a^2} = |a|$ (absolute value of $a$), so that your equation only holds if $a\geq 0$ to begin with. – Arturo Magidin Nov 04 '19 at 02:46
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The square root is a function, so it only takes one value per input (the positive, by convention). $$\sqrt{\cdot}: \mathbb{R}^+_0 \to \mathbb{R}_0^+ $$ $$x \mapsto \sqrt{x} $$ where $\sqrt{x}$ is the unique nonnegative real number such that $(\sqrt{x})^2 = x$.

It can also be proven that, if $a$ and $b$ are nonnegative, it holds that $a = b \Leftrightarrow \sqrt{a}=\sqrt{b}$. (Edit: This essentially means that the square root function is one-to-one.)

Solving these equations relies on this fact. The first one is straightforward: $x = 9 \Leftrightarrow \sqrt{x} = \sqrt{9} = 3$ (you cannot write $\sqrt{x} = - 3$, because by definition the square root only takes nonnegative values).

The second equation seems to contradict this, because of the two solutions, but if you look closely at what's happening, it becomes clearer: $$a^2 = 9 \Leftrightarrow \sqrt{a^2} = \sqrt{9} \Leftrightarrow |a| = 3 \Leftrightarrow a = -3 \:\lor a=3$$.

The catch here is that, because of the nonnegative value condition, $\sqrt{x^2} = |x| \neq x$. Note that the two solutions don't come from taking the square root, but actually from taking the absolute value!

Many teachers skip that intermediate step, so it ends up looking like the square root is the cause of the two solutions, even though it isn't.

J. C.
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You are right that often $\sqrt{n}$ is meant as the positive root. As a result, it is very common to see the solutions to $x^2=n$ expressed as $x=\pm\sqrt{n}$ rather than $\pm x=\sqrt{n}$. While they are equivalent mathematically, the latte

Shon Verch
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