# How do you simplify sqrt(9e^6)?

Aug 21, 2016

It depends what you mean by $e$...

#### Explanation:

• If $e$ is the mathematical constant ($\approx 2.7182818$), then the answer is definitely $3 {e}^{3}$.

• If $e$ is a Real valued variable, then $3 \left\mid {e}^{3} \right\mid$ covers both positive and negative values of $e$.

• If $e$ is a Complex valued variable then $3 \sqrt{{e}^{6}}$ is about the best you can do.

$\textcolor{w h i t e}{}$
The square root symbol sqrt denotes the principal square root.

We have:

${\left(3 {e}^{3}\right)}^{2} = {3}^{2} {\left({e}^{3}\right)}^{2} = 9 {e}^{6}$

${\left(- 3 {e}^{3}\right)}^{2} = {\left(- 3\right)}^{2} {\left({e}^{3}\right)}^{2} = 9 {e}^{6}$

So regardless of the value of $e$ (Real or Complex), both $3 {e}^{3}$ and $- 3 {e}^{3}$ are square roots of $9 {e}^{6}$, but which is the principal one?

If $3 {e}^{3}$ is a non-negative Real number then it is the principal square root. Hence for $e \ge 0$ we have $\sqrt{9 {e}^{6}} = 3 {e}^{3}$

If $3 {e}^{3}$ is a negative Real number then $- 3 {e}^{3}$ is positive and $\sqrt{9 {e}^{6}} = - 3 {e}^{3}$.

So if $e$ is a Real number then $\sqrt{9 {e}^{6}} = \left\mid 3 {e}^{3} \right\mid = 3 \left\mid {e}^{3} \right\mid$

$\textcolor{w h i t e}{}$

Suppose $e = \sqrt{i} = \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)$

Then:

$3 {e}^{3} = 3 \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

$\sqrt{9 {e}^{6}} = 3 \sqrt{\cos \left(\frac{6 \pi}{4}\right) + i \sin \left(\frac{6 \pi}{4}\right)}$

$= 3 \sqrt{\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)}$

$= 3 \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

$= - 3 \left(\cos \left(\frac{3 \pi}{4}\right) + \sin \left(\frac{3 \pi}{4}\right)\right)$

$= - 3 {e}^{3}$

So in the general Complex case, about the best we can say is:

$\sqrt{9 {e}^{6}} = \pm 3 {e}^{3}$

or better:

$\sqrt{9 {e}^{6}} = 3 \sqrt{{e}^{6}}$

It would be possible to pick out individual cases according to their $A r g$ value, but it would hardly be simplifying.