Question

Can someone help me find the 5 solutions to this equation?

Answer 1

\[y ^{5}-100y=0\]

Answer 2

y=0 is one

Answer 3

use de moivre's theorem

Answer 4

I don't know what that is. could you explain please?

Answer 5

-sqrt(10)
0
sqrt(10)
y = -i sqrt(10)
y = i sqrt(10)

Answer 6

Do you know euler's formula?

Answer 7

where e^(i*pi) = -1

Answer 8

yeah, I think I learned a bit about that. I don't remember these names though. I don't think my teacher did a good job explaining. thank you

Answer 9

and e^(ix) = cosx+isin(x)

Answer 10

to clarify, is this what the answer should look like, or is it in a different format?\[\sqrt[i]{10}\]

Answer 11

yes

Answer 12

What class are you in?

Answer 13

college algebra

Answer 14

1 or 2 ? because if you are in college algebra, you would use synethic division to solve these things.

Answer 15

Euler's formula is like pre-calc.. and trig-ish

Answer 16

1

Answer 17

I think we are just using substitution to solve for thse. thats all it really says in this chapter of my book

Answer 18

I think most you need to solve this problem is the ability to factorize the expression, mainly difference of two squares.

Answer 19

so will i be getting different answers than was given?

Answer 20

No, the answers given are right! But do you know how to get them?

Answer 21

not exactly

Answer 22

Well, first take y as a common factor, you get:
\[y(y^4-100)=0\]
Now, y^4-100 is a difference of two squares. It can factorized using the formula:
a^2-b^2=(a-b)(a+b)

Answer 23

Ok
what you need to do is use factor

After you factor to Y (y^2-10) (y^2+10)
You can use quadratic formula on the last term (y^2+10) to get the imaginary roots.

Answer 24

Applying this formula to our equation gives:
\[y(y^2-10)(y^2+10)=0\]

Answer 25

Are you following so far? If something does not make sense to you, let me know before we proceed.

Answer 26

it looks good so far. Thanks

Answer 27

Good. Now you have:
\[y=0\]
\[y^2-10=0 \implies y^2=10 \implies y=\sqrt{10}, y=-\sqrt{10}\]
These are three solutions.

Answer 28

We still have two, which are:
\[y^2+10=0 \implies y^2=-10\]
What do you think about this last equation?

Answer 29

hmm. That sort of makes sense.Sorry, i'm not exactly what you mean by what do I think about the last equation?

Answer 30

OK. You can see that we have y^2=(-)10, while square of a real number is always a positive value. In this case y is a complex number. That's:
\[y^2=-10 \implies y=\sqrt{10}i, y=-\sqrt{10}i\]
The set of solutions to the given equation is:
\[\left\{ 0,\sqrt{10},-\sqrt{10},\sqrt{10}i.-\sqrt{10}i \right\}\]

Answer 31

It's a comma between the last two numbers, not a point.

Answer 32

oh ok. Thank you for your help

Answer 33

You're welcome!