# Root help?!

## If $c = {d}^{2}$, does $d = \sqrt{c}$?

Jun 8, 2017

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 the principal square root
• a negative square root

That is the case because the square root of a positive real number $c$, let's say $d$ to use the variables you have in your example, is defined as the number that, if multiplied by itself, gives you $d$.

In other words, if you have

$d \times d = {d}^{2} = c$

then you can say that

$d = \sqrt{c}$

is the square root of $c$.

However, notice what happens if we multiply $- d$ by itself

$\left(- d\right) \times \left(- d\right) = \left(d \times d\right) = {d}^{2} = c$

This time, you can say that

$d = - \sqrt{c}$

is the square root of $c$.

Therefore, for every positive real number $c$, you have two possible square roots denoted using a plus-minus sign

$d = \pm \sqrt{c}$

You can thus say that if

$c = {d}^{2}$

then

$d = \pm \sqrt{c}$

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

${d}^{2} = {\left(+ \sqrt{c}\right)}^{2} \text{ }$ and $\text{ } {d}^{2} = {\left(- \sqrt{c}\right)}^{2}$

which is

${d}^{2} = \sqrt{c} \cdot \sqrt{c} \text{ }$ and $\text{ } {d}^{2} = \left(- \sqrt{c}\right) \cdot \left(- \sqrt{c}\right)$

${d}^{2} = \sqrt{c} \cdot \sqrt{c} \text{ }$ and $\text{ } {d}^{2} = \sqrt{c} \cdot \sqrt{c}$

${d}^{2} = c \text{ }$ and $\text{ } {d}^{2} = c$

So, for example, you can say that the square roots of $25$ are

$\sqrt{25} = \pm 5$

The principal square root of $25$ is equal to $5$, which is why we always say that

$\sqrt{25} = 5$

but do not forget that $- 5$ is also a square root for $25$, since

$\left(- 5\right) \cdot \left(- 5\right) = 5 \cdot 5 = {5}^{2} = 25$