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Math Help - Help solving for x please... 2(x^2 -1)^(1/5)

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    Help solving for x please... 2(x^2 -1)^(1/5)

    Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

    <br />
f(x) = 2(x^2 -1)^\frac{1}{5}<br />

    Edit: Oops, yeah, I want to solve for f(x) = 0. I'm trying to find the x-intercepts.
    Last edited by liquidwater; April 1st 2010 at 10:45 PM.
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    Quote Originally Posted by liquidwater View Post
    Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

    <br />
f(x) = 2(x^2 -1)^\frac{1}{5}<br />
    Dear liquidwater,

    Do you mean you want to subject x in this equation???
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    Quote Originally Posted by liquidwater View Post
    Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

    <br />
f(x) = 2(x^2 -1)^\frac{1}{5}<br />
    You need an equation in order to solve for x. Is this equal to 0?


    Anyway, if you want to factorise it...

    f(x) = 2(x^2 - 1)^{\frac{1}{5}}

     = 2[(x - 1)(x + 1)]^{\frac{1}{5}}

     = 2(x - 1)^{\frac{1}{5}}(x + 1)^{\frac{1}{5}}.
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    Quote Originally Posted by liquidwater View Post
    Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

    <br />
f(x) = 2(x^2 -1)^\frac{1}{5}<br />
    Dear liquidwater,

    Do you mean you want to subject x in this equation???
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    Quote Originally Posted by Sudharaka View Post
    Dear liquidwater,

    Do you mean you want to subject x in this equation???
    Thanks, I updated my original post.

    Prove It, thanks... I still can't figure how to find f(x) = 0 with that.
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    2(x^2 - 1)^{\frac{1}{5}} = 0

    2[(x - 1)(x + 1)]^{\frac{1}{5}} = 0

    2(x - 1)^{\frac{1}{5}}(x + 1)^{\frac{1}{5}} = 0.


    So (x - 1)^{\frac{1}{5}} = 0 or (x + 1)^{\frac{1}{5}} = 0.


    Case 1:

    (x - 1)^{\frac{1}{5}} = 0

    x - 1 = 0

    x = 1.


    Case 2:

    (x + 1)^{\frac{1}{5}} = 0

    x + 1 = 0

    x = -1.
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    Quote Originally Posted by Prove It View Post
    2(x^2 - 1)^{\frac{1}{5}} = 0

    2[(x - 1)(x + 1)]^{\frac{1}{5}} = 0

    2(x - 1)^{\frac{1}{5}}(x + 1)^{\frac{1}{5}} = 0.


    So (x - 1)^{\frac{1}{5}} = 0 or (x + 1)^{\frac{1}{5}} = 0.


    Case 1:

    (x - 1)^{\frac{1}{5}} = 0

    x - 1 = 0

    x = 1.


    Case 2:

    (x + 1)^{\frac{1}{5}} = 0

    x + 1 = 0

    x = -1.
    Thanks a lot.

    Can the ^\frac{1}{5} be discarded because we will be making x = 0, 0 to the power of anything is just going to be 0?

    And 2*0 is just going to be 0... so the 2 is discarded?

    (if I'm making sense).
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    Quote Originally Posted by liquidwater View Post
    Thanks a lot.

    Can the ^\frac{1}{5} be discarded because we will be making x = 0, 0 to the power of anything is just going to be 0?

    And 2*0 is just going to be 0... so the 2 is discarded?

    (if I'm making sense).
    In this case you could, but if the RHS wasn't 0, you'd have to take both sides to the power of 5.
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    Quote Originally Posted by Prove It View Post
    In this case you could, but if the RHS wasn't 0, you'd have to take both sides to the power of 5.
    Thanks so much, I understand
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  10. #10
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    You can, of course, take the fifth power here:

    2(x^2- 1)^{1/5}= 0

    Divide both sides by 2: (x^2- 1)^{1/5}= 0/2= 0

    Take the fifth power of both sides: x^2- 1= 0

    Add 1 to both sides: x^2= 1

    So x= \pm 1.

    More generally, if 2(x^2- 1)^{1/5}= a,
    for a any number, (x^2- 1)^{1/5}= \frac{a}{2}
    x^2- 1= \left(\frac{a}{2}\right)^5= \frac{a^5}{32}
    x^2= 1+ \frac{a^5}{32}

    x= \pm\sqrt{1+ \frac{a^5}{32}}
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