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

- anonymous

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

- jamiebookeater

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- anonymous

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

- anonymous

y=0 is one

- anonymous

use de moivre's theorem

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## More answers

- anonymous

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

- anonymous

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

- anonymous

Do you know euler's formula?

- anonymous

where e^(i*pi) = -1

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

What class are you in?

- anonymous

college algebra

- anonymous

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

- anonymous

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

- anonymous

1

- anonymous

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

- anonymous

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

- anonymous

so will i be getting different answers than was given?

- anonymous

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

- anonymous

not exactly

- anonymous

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)

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

it looks good so far. Thanks

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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:
\[\left\{ 0,\sqrt{10},-\sqrt{10},\sqrt{10}i.-\sqrt{10}i \right\}\]

- anonymous

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

- anonymous

oh ok. Thank you for your help

- anonymous

You're welcome!

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