Question

Solve the equation. (Sort your answers in increasing order, first by their imaginary parts, then by their real parts.)
x^4 + 81 = 0

Answer 1

you could factor x^4+81
\[a^2+b^2=(a+bi)(a-bi)\]

Answer 2

but it look like you have been asking trig questions so i have a better way

Answer 3

this might seem like a stupid question but that is only for squared... what abt the rest of the problem?

Answer 4

or would it be twice?

Answer 5

\[x^4=-81 \\ x^4=-81(\cos(0)+i \sin(0)) \\ \text{ so since cos and sin are periodic we have } \\x^4=-81(\cos(0+2n \pi)+i \sin(0+2n \pi))\]
you want 4 answers
so find your first answer when n=0
next answer when n=1
next n=2
next n=3
that is 4 roots

Answer 6

\[x^4+9^2+18x^2-18x^2=0\]
\[\left( x^2+9 \right)^2-\left( 3\sqrt{2}x \right)^2=0\]
\[\left( x^2+3\sqrt{2}x+9 \right)\left( x^2-3\sqrt{2}x+9 \right)=0\]
solve the two quadratics separately by quadratic formula.

Answer 7

cos and sin have period 2pi that is

Answer 8

and you need to take the 4th root of both sides too :)

Answer 9

\[x^4=-81 \\ x^4=81(\cos(\pi)+i \sin(\pi)) \\ \text{ so since \cos and \sin are periodic we have } \\x^4=81(\cos(\pi+2n \pi)+i \sin(\pi+2n \pi)) \]
that -81 didn't help much

Answer 10

geeze. so which strategy do I go with, yours or surj?

Answer 11

either

Answer 12

ok let me start figuring out

Answer 13

yay! it worked i went your way, it just seemed easier to me since I am already familiar with it

Answer 14

ok and I changed it from cos(0)+i sin(0) to cos(pi)+isin(pi)
because I didn't want to take the square root of -81

Answer 15

well it worked thank u...

Answer 16

ok

Answer 17

ooops 4th root whatever lol