1. Difficult algebra..

How can I isolate y? y>0

55(1-y^20) = (1 - y^10)[30 + 60y^10 + 90y^20]

I'm having a really tough time. I get to:
0 = -3y^30 + 2.83 y^20 + y^10 - 0.83

I have no idea how to solve this kind of equation. Help?
(I realize I could replace y^10 = x, say, and get a third degree polynomial, but I'm not in the measure of solving such an equation ...)

2. Originally Posted by Volcanicrain
How can I isolate y? y>0

55(1-y^20) = (1 - y^10)[30 + 60y^10 + 90y^20]

I'm having a really tough time. I get to:
0 = -3y^30 + 2.83 y^20 + y^10 - 0.83

I have no idea how to solve this kind of equation. Help?
(I realize I could replace y^10 = x, say, and get a third degree polynomial, but I'm not in the measure of solving such an equation ...)
Notice that you can factor the left side as the difference of two squares:

$55(1-y^{20})=55(1-y^{10})(1+y^{10})$

Do you see how you can cancel now?

3. Wow that's amazing! Thanks a million You really helped me out!

4. You can now write:

$55(1+y^{10})=30+60y^{10}+90y^{20}$

If you let $u=10$ you can write this as:

$55(1+y^{u})=30+60y^{u}+90y^{2u}$

The let $c=1^u$

So now this takes the form of a quadratic.

$55(1+cy)=30+60cy+90cy^2$

Where $c$ is can be treated as a constant. You should be able to isolate y by factoring or using the quadratic formula.

5. I was able to, thanks a lot I got the right answer for my problem!!
You're a godsend!
Could I directly substitute k = y^10 ?

6. Originally Posted by Volcanicrain
I was able to, thanks a lot I got the right answer for my problem!!
You're a godsend!
Could I directly substitute k = y^10 ?
Absolutely