Posted by **Hellga** on Wednesday, January 9, 2008 at 8:55pm.

Please solve this equation. As you write down the steps, please give me a description on what you did.

y^3 - 16y^3/2 + 64 = 0

- Trig. -
**Damon**, Wednesday, January 9, 2008 at 8:56pm
I tried to answer down below

I think no real y is a solution

- Trig. -
**Damon**, Wednesday, January 9, 2008 at 8:58pm
OH, that is a different problem !

let p = y^1.5

then

p^2 -16 p + 64 = 0

- Trig. -
**Hellga**, Wednesday, January 9, 2008 at 8:59pm
Explain!!!!!!

- Trig. -
**Damon**, Wednesday, January 9, 2008 at 9:01pm
(p-8)(p-8) = 0

p = 8

y^3/2 = 8

y^3 = 8*8

y = 2*2 = 4

- Trig. -
**Damon**, Wednesday, January 9, 2008 at 9:02pm
The one with the typo was hard.

- Trig. -
**Damon**, Wednesday, January 9, 2008 at 9:05pm
Explain:

I saw (once you fixed it) that the variable at the beginning, y^3 was the square of the one in the middle y^3/2

So I figured I could make it into a quadratic if I called the 3/2 one p and then the 3 one would be p^2

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