# Square root of number: concept problem [duplicate]

I know that the square root of $9$ is $3$

But somebody told me that

If $x^2= 9$ we get the solution as $x=\pm3$

I am confused when do we have $\sqrt{9}=\pm3$ and only $3$?

## marked as duplicate by Dietrich Burde, Clayton, Hans Lundmark, Jendrik Stelzner, Lord Shark the UnknownAug 10 '18 at 5:23

• I do not agree that this question should be closed. – Kevin Aug 9 '18 at 14:48
• This is simply a definition. Any positive number $x$ has two square roots. The symbol $\sqrt{x}$ refers to the positive value, by definition. – saulspatz Aug 9 '18 at 14:59
• @Kevin: This has been asked many times before. For example, look at the list of linked question to the other duplicate that I suggested: math.stackexchange.com/questions/linked/26363?lq=1 – Hans Lundmark Aug 9 '18 at 15:06
• @HansLundmark Ah, then if it a matter of house rules and the question is repettitve then fair enough. I had thought the closure was down to the quality of the post. Thank you. – Kevin Aug 10 '18 at 9:15

We do not "have $\sqrt{9}=\pm3$".

We have only $\sqrt9 = 3.$

The expression $\sqrt9$ can never have the value $-3.$ Never, never, never. (OK, maybe in some relatively obscure branch of math there is such a notation. But not in the standard real arithmetic people generally learn in high school.)

What can happen is that we are looking for the solutions of $x^2 = 9.$ This equation has solutions for two values of $x.$ One of these solutions is $x = \sqrt9.$ The other solution is $x = -\sqrt9.$

Since $\sqrt9 = 3$ (always!), it follows that $$-\sqrt9 = -3,$$ and that's how you get the negative solution of $x^2 = 9.$

Sometimes, someone might write that the solutions of $x^2 = 9$ are $x = \pm\sqrt9,$ and you could further write $\pm\sqrt9 = \pm 3.$ But notice that there are $\pm$ signs on both sides of that equation.

Take any positive number. For ease, call it $a$. Then it's square is $a^2$.

If we take $a$ and look on the number line to $-a$, you find that this square is also $a^2$. This is because $(-a)^2=(-1)^2 \cdot a^2 = (-1)\cdot(-1)\cdot a^2=a^2$

This means that square rooting any number can hove two possible values, in this case those two values can either be $a$ or $-a$.

When you come to solving equations (I don't know \ can't tell your current working level) you will see that my example is similar to solving $x^2-a^2=0$.

When solving equation it may given some information about the equation, be able to tell whether you choose the positive root or the negative one.

$x^2=9$

$x=\pm\sqrt{9}$ ( We took square root of both sides of the equation)

$x=\pm3$

So if we isolate the part $\sqrt{9}$ it is equal to $3$

We start with $x^2=9$, we take square root of both sides and we get $\sqrt{x^2}=\sqrt{9}$ or $|x|=3$. The last equation has two solutions but $\sqrt{9}=3$. There is no confusion about that.