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Math Help - Algebra 2 Super Challenge Problem!

  1. #1
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    Talking Algebra 2 Super Challenge Problem!

    Hello everyone, this is my first post in the Math Help Forum. I was wanting to know if anybody out there can prove that x = 3 (without just 'plugging it' in) in this equation that my friend made:

    3√(8x+3) - 3√(x-4) = 4 (third roots)

    Here's what I have of the proof so far: View image: Proof of the Super Equation( WIP) Sorry if you can't follow it lol...
    What would you recommend I do as the next step? The numbers are really getting out of control...
    Last edited by odysseyofnoises; November 4th 2012 at 11:17 PM.
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  2. #2
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    Re: Algebra 2 Super Challenge Problem!

    At the point where you have 8x+3 = 64 + ..... , rearrange the equation as

    7(x-4) - 12(x-4)^{2/3}-48(x-4)^{1/3} - 29 = 0.

    Now make a substitution, z = (x-4)^{1/3}, and you arrive at the cubic equation

    7z^{3}-12z^{2}-48z-29 = 0.

    Whether or not you accept trial values for z (which is a normal method) I don't know. ( z = -1 is a fairly obvious solution.) The alternative is to use perhaps Cardano's method.
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  3. #3
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    Re: Algebra 2 Super Challenge Problem!

    How did you go from the 8x+3 = 64... to the 7(x-4) one? I don't see how the flow of logic works from one step to the next. I understand the substitution part though. I got this: View image: Proof of the Super Equation( WIP)
    Last edited by odysseyofnoises; November 5th 2012 at 05:03 PM.
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  4. #4
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    Re: Algebra 2 Super Challenge Problem!

    -7x+57 = -7x+28+29=-7(x-4)+29.
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    Re: Algebra 2 Super Challenge Problem!

    Oh, that makes sense :P you just rewrite those into an (x-4) term so that it can be replaced with substitution. Now I'll just work with the cubic function soon (maybe not today, have too much homework...). And yes, I accept trial values for z, otherwise I would probably use some extremely complicated method...
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    Re: Algebra 2 Super Challenge Problem!

    Now, I am doing polynomial division between (z+1) and the cubic function. I am getting a remainder, what course of action do you recommend now? here's what I have so far: View image: Proof of the Super Equation( WIP)
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  7. #7
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    Re: Algebra 2 Super Challenge Problem!

    You should not get a remainder in the polynomial long division. Doing the division, I find:

    7z^3-12z^2-48z-29=(z+1)(7z^2-19z-29)
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    Re: Algebra 2 Super Challenge Problem!

    I think I see what I did wrong here... bad arithmetic. Oh well, someday I'll learn to count :P
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    Re: Algebra 2 Super Challenge Problem!

    Quote Originally Posted by BobP View Post
    At the point where you have 8x+3 = 64 + ..... , rearrange the equation as

    7(x-4) - 12(x-4)^{2/3}-48(x-4)^{1/3} - 29 = 0.

    Now make a substitution, z = (x-4)^{1/3}, and you arrive at the cubic equation

    7z^{3}-12z^{2}-48z-29 = 0.

    Whether or not you accept trial values for z (which is a normal method) I don't know. ( z = -1 is a fairly obvious solution.) The alternative is to use perhaps Cardano's method.
    For the record the cubic has the 3 real solutions as
    z = -1,~\frac{19 \pm \sqrt{1173}}{14}

    So
    x = 3,~\frac{10587 \pm 282\sqrt{1173}}{343}

    All of these are correct according to the original equation, but I would highly recommend the last two solutions be done on a calculator. (You'll see why when you plug them into the original equation.)

    Okay, I was bored.

    -Dan
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