# What is the square root of -8?

Jul 22, 2018

$\pm 2 i \sqrt{2}$

#### Explanation:

We have the following:

$\sqrt{- 8}$

We can rewrite this as

$\textcolor{s t e e l b l u e}{\sqrt{8}} \cdot \textcolor{p u r p \le}{\sqrt{- 1}}$

Recall that $i = \sqrt{- 1}$

We can further break this down as

$\textcolor{s t e e l b l u e}{\sqrt{4} \cdot \sqrt{2}} \cdot \textcolor{p u r p \le}{i}$, or

$\pm 2 i \sqrt{2}$

Hope this helps!

$\setminus \sqrt{- 8} = \setminus \pm 2 i \setminus \sqrt{2}$

#### Explanation:

The square root of $- 8$ is given

$\setminus \sqrt{- 8}$

$= {\left(- 8\right)}^{\frac{1}{2}}$

$= {\left(8 \left(\setminus \cos \setminus \pi + i \setminus \sin \setminus \pi\right)\right)}^{\frac{1}{2}}$

$= {8}^{\frac{1}{2}} {\left(\setminus \cos \setminus \pi + i \setminus \sin \setminus \pi\right)}^{\frac{1}{2}}$

$= 2 \setminus \sqrt{2} {\left(\setminus \cos \left(2 k \setminus \pi + \setminus \pi\right) + i \setminus \sin \left(2 k \setminus \pi + \setminus \pi\right)\right)}^{\frac{1}{2}}$

$= 2 \setminus \sqrt{2} {\left(\setminus \cos \left(\left(2 k + 1\right) \setminus \pi\right) + i \setminus \sin \left(\left(2 k + 1\right) \setminus \pi\right)\right)}^{\frac{1}{2}}$

$= 2 \setminus \sqrt{2} \left(\setminus \cos \left(\frac{\left(2 k + 1\right) \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{\left(2 k + 1\right) \setminus \pi}{2}\right)\right)$

Where, $k = 0 , 1$

Now, setting $k = 0$ & $k = 1$, we get two values as follows

$\setminus \sqrt{- 8} = 2 \setminus \sqrt{2} \left(\setminus \cos \left(\frac{\left(2 \left(0\right) + 1\right) \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{\left(2 \left(0\right) + 1\right) \setminus \pi}{2}\right)\right)$

$\setminus \sqrt{- 8} = 2 i \setminus \sqrt{2}$ &

$\setminus \sqrt{- 8} = 2 \setminus \sqrt{2} \left(\setminus \cos \left(\frac{\left(2 \left(1\right) + 1\right) \setminus \pi}{2}\right) + i \setminus \sin \left(\frac{\left(2 \left(1\right) + 1\right) \setminus \pi}{2}\right)\right)$

$\setminus \sqrt{- 8} = - 2 i \setminus \sqrt{2}$

$\setminus \therefore \setminus \sqrt{- 8} = \setminus \pm 2 i \setminus \sqrt{2}$

Jul 22, 2018

color(green)(sqrt(-8)=2sqrt2i or sqrt(-8)=-2sqrt2i

#### Explanation:

Let ,

$x = \sqrt{- 8} \ldots \to x \notin \mathbb{R} , b u t , x \in \mathbb{C}$

Squaring both sides

${x}^{2} = - 8 = 8 \left(- 1\right)$

$\therefore {x}^{2} = 8 {i}^{2} \to \left[\because {i}^{2} = - 1\right]$

$\therefore {x}^{2} - 8 {i}^{2} = 0$

$\therefore {x}^{2} - {\left(2 \sqrt{2} i\right)}^{2} = 0$

:.(x-2sqrt2i)(x+2sqrt2i)=0to[color(blue)(because a^2- b^2=color(blue)((a-b)(a+b))]

$\therefore x - 2 \sqrt{2} i = 0 \mathmr{and} x + 2 \sqrt{2} i = 0$

$\therefore x = 2 \sqrt{2} i \mathmr{and} x = - 2 \sqrt{2} i$

:.color(green)(sqrt(-8)=2sqrt2i or sqrt(-8)=-2sqrt2i
...................................................................................

Note: If $x = \sqrt{8} = \sqrt{4 \times 2} = \sqrt{{\left(2 \sqrt{2}\right)}^{2}}$

:.color(red)(x=sqrt8=+-2sqrt2 inRR