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November 29, 2014

November 29, 2014

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

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

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

I think no real y is a solution

- Trig. -
**Hellga**, Wednesday, January 9, 2008 at 8:57pmIt's a different problem...I changed one of the terms..Try again!

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

let p = y^1.5

then

p^2 -16 p + 64 = 0

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

- Trig. -
- 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 9:02pmThe one with the typo was hard.

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

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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