# Root help?!

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If #c = d^2# , does #d = sqrt c# ?

If

##### 1 Answer

#### Answer:

Yes, but that's only half the story.

#### Explanation:

The thing to remember here is that every *positive* real number has **two square roots**

a positive square root called theprincipal square roota negative square root

That is the case because the square root of a positive real number **itself**, gives you

In other words, if you have

#d xx d = d^2 = c#

then you can say that

#d = sqrt(c)#

is the square root of

However, notice what happens if we multiply

#(-d) xx (-d) = (d xx d) = d^2 = c#

This time, you can say that

#d = -sqrt(c)#

is the square root of

Therefore, for every positive real number **two possible square roots** denoted using a plus-minus sign

#d = +- sqrt(c)#

You can thus say that if

#c = d^2#

then

#d = +- sqrt(c)#

You can check that this is the case because if you square both side, you will end up with

#d^2 = (+sqrt(c))^2" "# and#" "d^2 = (-sqrt(c))^2#

which is

#d^2 = sqrt(c) * sqrt(c)" "# and#" " d^2 = (-sqrt(c)) * (-sqrt(c))#

#d^2 = sqrt(c) * sqrt(c)" "# and#" " d^2 = sqrt(c) * sqrt(c)#

#d^2 = c" "# and# " "d^2 = c#

So, for example, you can say that the square roots of

#sqrt(25) = +-5#

The **principal square root** of

#sqrt(25) = 5#

but do not forget that

#(-5) * (-5) = 5 * 5 = 5^2 = 25#