## anonymous 5 years ago Can someone help me find the 5 solutions to this equation?

1. anonymous

$y ^{5}-100y=0$

2. anonymous

y=0 is one

3. anonymous

use de moivre's theorem

4. anonymous

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

5. anonymous

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

6. anonymous

Do you know euler's formula?

7. anonymous

where e^(i*pi) = -1

8. 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

9. anonymous

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

10. anonymous

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

11. anonymous

yes

12. anonymous

What class are you in?

13. anonymous

college algebra

14. anonymous

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

15. anonymous

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

16. anonymous

1

17. anonymous

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

18. anonymous

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

19. anonymous

so will i be getting different answers than was given?

20. anonymous

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

21. anonymous

not exactly

22. 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)

23. 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.

24. anonymous

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

25. anonymous

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

26. anonymous

it looks good so far. Thanks

27. 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.

28. anonymous

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

29. anonymous

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

30. 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: $\left\{ 0,\sqrt{10},-\sqrt{10},\sqrt{10}i.-\sqrt{10}i \right\}$

31. anonymous

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

32. anonymous

oh ok. Thank you for your help

33. anonymous

You're welcome!