Question

What is the square root of -8?

Answer 1

`pm2isqrt2`

Explanation:

We have the following:

`sqrt(-8)`

We can rewrite this as

`color(steelblue)(sqrt8)*color(purple)(sqrt(-1))`

Recall that `i=sqrt(-1)`

We can further break this down as

`color(steelblue)(sqrt(4)*sqrt2)*color(purple)i`, or

`pm2isqrt2`

Hope this helps!

Answer 2

`\sqrt{-8}=\pm 2i\sqrt2`

Explanation:

The square root of `-8` is given

`\sqrt{-8}`

`=(-8)^{1/2}`

`=(8(\cos\pi+i\sin\pi))^{1/2}`

`=8^{1/2}(\cos\pi+i\sin\pi)^{1/2}`

`=2\sqrt2(\cos(2k\pi+\pi)+i\sin(2k\pi+\pi))^{1/2}`

`=2\sqrt2(\cos((2k+1)\pi)+i\sin((2k+1)\pi))^{1/2}`

`=2\sqrt2(\cos({(2k+1)\pi}/2)+i\sin({(2k+1)\pi}/2))`

Where, `k=0, 1`

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

`\sqrt{-8}=2\sqrt2(\cos({(2(0)+1)\pi}/2)+i\sin({(2(0)+1)\pi}/2))`

`\sqrt{-8}=2i\sqrt2` &

`\sqrt{-8}=2\sqrt2(\cos({(2(1)+1)\pi}/2)+i\sin({(2(1)+1)\pi}/2))`

`\sqrt{-8}=-2i\sqrt2`

`\therefore \sqrt{-8}=\pm 2i\sqrt2`

Answer 3

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

Explanation:

Let ,

`x=sqrt(-8)...tox!inRR, but , x in CC`

Squaring both sides

`x^2=-8=8(-1)`

`:.x^2=8i^2 to[because i^2=-1]`

`:.x^2-8i^2=0`

`:.x^2-(2sqrt2 i)^2=0`

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

`:.x-2sqrt2i=0 or x+2sqrt2i=0`

`:.x=2sqrt2i or x=-2sqrt2i`

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

Note: If `x=sqrt8=sqrt(4 xx2)=sqrt((2sqrt2)^2)`

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