# Question 586a5

Aug 30, 2017

#### Explanation:

Neither $3 \sqrt{- 64}$ or $\sqrt{- 64}$ can be simplified to real numbers.
The square root of a negative number is always an imaginary number. The examples can be simplified to have the lowest value radicand, but this will still be an imaginary number.

From examples:

$3 \sqrt{- 64} \implies$ $3 \sqrt{64 \cdot \left(- 1\right)}$$\implies$ $24 \sqrt{- 1}$

$\sqrt{- 64} \implies$ sqrt(64*(-1)# $\implies 8 \sqrt{- 1}$

These can also be expressed as $24 i$ and $8 i$ or $\left(0 + 24 i\right)$ and $\left(0 + 8 i\right)$, where $i$ is the imaginary unit $i =$ $\sqrt{- 1}$.
These are known as pure imaginary numbers.

Aug 30, 2017

Because a negative number raised to an odd power gives a negative result.

#### Explanation:

${\left(- 3\right)}^{3} = - 27$, so $\sqrt[3]{- 27} = - 3$

${\left(- 2\right)}^{5} = - 32$, so $\sqrt[5]{- 32} = - 2$

But for even powers, both positives and negatives raised to an even power give a positive result.

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

And any other number raised to the power $2$ will give a positive answer (except ${0}^{2} = 0$).

So no number can be raised to the power $2$ and give a negative result.

(Until we introduce a new kind of numbers that lets us do that.)