anonymous
  • anonymous
Can someone help me find the 5 solutions to this equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[y ^{5}-100y=0\]
anonymous
  • anonymous
y=0 is one
anonymous
  • anonymous
use de moivre's theorem

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anonymous
  • anonymous
I don't know what that is. could you explain please?
anonymous
  • anonymous
-sqrt(10) 0 sqrt(10) y = -i sqrt(10) y = i sqrt(10)
anonymous
  • anonymous
Do you know euler's formula?
anonymous
  • anonymous
where e^(i*pi) = -1
anonymous
  • 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
  • anonymous
and e^(ix) = cosx+isin(x)
anonymous
  • anonymous
to clarify, is this what the answer should look like, or is it in a different format?\[\sqrt[i]{10}\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
What class are you in?
anonymous
  • anonymous
college algebra
anonymous
  • anonymous
1 or 2 ? because if you are in college algebra, you would use synethic division to solve these things.
anonymous
  • anonymous
Euler's formula is like pre-calc.. and trig-ish
anonymous
  • anonymous
1
anonymous
  • anonymous
I think we are just using substitution to solve for thse. thats all it really says in this chapter of my book
anonymous
  • anonymous
I think most you need to solve this problem is the ability to factorize the expression, mainly difference of two squares.
anonymous
  • anonymous
so will i be getting different answers than was given?
anonymous
  • anonymous
No, the answers given are right! But do you know how to get them?
anonymous
  • anonymous
not exactly
anonymous
  • 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
  • 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
  • anonymous
Applying this formula to our equation gives: \[y(y^2-10)(y^2+10)=0\]
anonymous
  • anonymous
Are you following so far? If something does not make sense to you, let me know before we proceed.
anonymous
  • anonymous
it looks good so far. Thanks
anonymous
  • 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
  • anonymous
We still have two, which are: \[y^2+10=0 \implies y^2=-10\] What do you think about this last equation?
anonymous
  • 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
  • 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
  • anonymous
It's a comma between the last two numbers, not a point.
anonymous
  • anonymous
oh ok. Thank you for your help
anonymous
  • anonymous
You're welcome!

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