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Posted by 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. - , Wednesday, January 9, 2008 at 8:56pm

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

  • Trig. - , Wednesday, January 9, 2008 at 8:57pm

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

  • Trig. - , 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. - , Wednesday, January 9, 2008 at 8:59pm

    Explain!!!!!!

  • Trig. - , 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. - , Wednesday, January 9, 2008 at 9:02pm

    The one with the typo was hard.

  • Trig. - , 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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