# Trig.

posted by .

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. -

I tried to answer down below
I think no real y is a solution

• Trig. -

It's a different problem...I changed one of the terms..Try again!

• Trig. -

OH, that is a different problem !

let p = y^1.5
then
p^2 -16 p + 64 = 0

• Trig. -

Explain!!!!!!

• Trig. -

(p-8)(p-8) = 0
p = 8

y^3/2 = 8
y^3 = 8*8
y = 2*2 = 4

• Trig. -

The one with the typo was hard.

• Trig. -

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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