Question

How are you going to get the root of x^4 + 4?

Answer 1

help! me ishaan i dont get this

Answer 2

me either

Answer 3

\[\sqrt{x^4+4}\]?

Answer 4

by roots do you mean:
x^4+4=0 ?
You won't have any real roots.

Answer 5

Oh sorry yea.=0. How bout imaginary roots?

Answer 6

imaginary? lol

Answer 7

http://www.wolframalpha.com/input/?i=x^4%2B4%3D0

Answer 8

\[x = 4^{\frac{1}{4}}\]
\[x = 2^\frac{1}{2}\]

Answer 9

\[x =\sqrt2\]

Answer 10

How did you do that? Haha

Answer 11

shouldnt it be \[-4^{1/4}\]?

Answer 12

Lol sorry
fiddle is right I am trying to get the solution you will have to wait a little

Answer 13

Okay! Thanks! :)

Answer 14

\[x = (-1)^{\frac{1}{4}}*4^{\frac{1}{2}}\]

Answer 15

ishaan i have posted some songs on chat do u remember them?

Answer 16

\[x = (-1)^{\frac{1}{4}}* \sqrt2\]

Answer 17

Lol Okay karate let me check

Answer 18

Now all we need to do is to find \((-1)^{\frac{1}{4}}\)

Answer 19

\( -4^{\frac{1}{4}} \) in polar form is \( \sqrt{2}e^{\frac{i \pi} {4}}\)
or equivalently \( \sqrt{2} ( cos(\pi/4) + i sin(\pi/4)) \)
this simplifies to 1+i
the other 3 roots are -1+i, -1-i, 1-i

Answer 20

ah pellet i forgot the polar form

Answer 21

nice work phi

Answer 22

\[x^4+4=0\]
\[x^4+4x^2+4-4x^2=0\]
\[(x^2+2)^2-(2x)^2=0\]
\[(x^2-2x+2)(x^2+2x+2)=0\]
\[x^2-2x+2=0\quad\Rightarrow\quad x_{1,2}=\frac{2\pm\sqrt{4-8}}{2}=1\pm i\] or
\[x^2+2x+2=0\quad\Rightarrow\quad x_{3,4}=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm i\]

Answer 23

nice tricks up there nikvist ^^^