Question

how to solve r^8+8r^4+16=0?

Answer 1

(r^4+4)(r^4+4)=0

Answer 2

then?

Answer 3

r^4=-4
r= fourth roots of -4

Answer 4

Yes now we will have complex roots. In other words, are roots will be imaginary. Have learned complex numbers? Have you been introduced to imaginary roots?

Answer 5

yes,

Answer 6

Alright then you know that the imaginary number\[i =\sqrt{-1}\]Correct?

Answer 7

yes

Answer 8

I need all of 8 solutions.

Answer 9

So then then in general, the sum of two squares may be written as\[a ^{2}+b ^{2}= (a +bi)(a -bi)\]

Answer 10

Hence you can factor each factor further by this principle.

Answer 11

I factor and it repeats many time while I need 8 separate solutions for my differential equations. :(.

Answer 12

Right so you had\[(r ^{4}+4)(r ^{4}+4)=0\]
\[(r ^{2}+2i)(r ^{2}-2i)(r ^{2}+2i)(r ^{2}-2i)=0\]Do you understand thus far?

Answer 13

yes,

Answer 14

Now factor each of these factors and you will have 8 factors.

Answer 15

I will work on it, you still be here, right? I will show you my work and you will check it for me, right?

Answer 16

I'm here for now, I'll wait a while for you. I'll help someone else mean while. When you're ready, just ask for me.

Answer 17

thanks , but it is not the case of r^8 = -1, so I cannot apply e^(ipi) right?

Answer 18

right

Answer 19

thanks, I will tag you later.

Answer 20

OK

Answer 21

\[r^2+2i=0\\r^2=-2i\\r=\pm\sqrt{2i}\]and it repeats with r^2 -2i =0. HA!! I don't get.

Answer 22

For \[r ^{2}-2i =0\]
\[r ^{2}=2i\]
\[r =\pm \sqrt{2i}\]

Answer 23

For \[r ^{2}+2i =0\]
\[r ^{2}=-2i\]
\[r =\pm \sqrt{-2i}\]If you wish to simplify further,\[r =\pm i \sqrt{2i}\]
\[r =\pm i ^{\frac{ 3 }{ 2 }}\sqrt{2}\]

Answer 24

your avatar looks exactly tcarroll010's one. At the first see, I tried to send message to you but cannot since you set your profile just receive message from person you fanned. I got mad with tcarroll010. I thought 'what is he doing?" hihihi.... when figure out the misunderstanding, I sent his message to let him know you "stole" his avatar.

Answer 25

but just 4 of them, not 8

Answer 26

I think he stole mine. Because I have had the same one for two years now.

Answer 27

Right, four equal and another four equal. For a total of eight.

Answer 28

I joined the site almost 1year, I didn't see you. Ha!!

Answer 29

Just like (x - 3)(x - 3) = 0 has two equal and real roots.

Answer 30

As a teacher, I'm busy, and thus not always available.

Answer 31

What grade are you in?

Answer 32

ok, got you. so, they are repeated and the solution for differential equations differentiated by multiply with polynomial. Thank you very much