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
p^2 -16 p + 64 = 0
Trig. - Hellga, Wednesday, January 9, 2008 at 8:59pm
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
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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