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Math Help - Interesting Equation! Solve for x.

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    Interesting Equation! Solve for x.

    Solve for x.

    x^3+1=2\sqrt[3]{2x-1}
    Last edited by thevinh; June 29th 2012 at 09:36 AM.
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    Re: Interesting Equation! Solve for x.

    One obvious solution is x= 1.
    There is one other positive root between 0 and 1 but it is not a rational number. There are either 2 or 4 negative roots, again irrational. There are either 0 or 2 complex conjugate roots.
    After that, I don't see anything reasonable.
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    Re: Interesting Equation! Solve for x.

    Quote Originally Posted by HallsofIvy View Post
    One obvious solution is x= 1.
    There is one other positive root between 0 and 1 but it is not a rational number. There are either 2 or 4 negative roots, again irrational. There are either 0 or 2 complex conjugate roots.
    After that, I don't see anything reasonable.
    This is answered in full in another thread.
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    Re: Interesting Equation! Solve for x.

    There's not really an easy algebraic solution. If you cube both sides, you get an ugly 9th degree polynomial. Sure, x = 1 is a root, but there are several others.
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    Re: Interesting Equation! Solve for x.

    Quote Originally Posted by richard1234 View Post
    There's not really an easy algebraic solution. If you cube both sides, you get an ugly 9th degree polynomial. Sure, x = 1 is a root, but there are several others.
    Actually there are only two more real solutions...
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    Re: Interesting Equation! Solve for x.

    Quote Originally Posted by richard1234 View Post
    There's not really an easy algebraic solution. If you cube both sides, you get an ugly 9th degree polynomial. Sure, x = 1 is a root, but there are several others.
    Actually there is an algebraic way to solve this problem without having to cube both sides. See solution posted at Give me maths problems!. Squaring or cubing to eliminate radicals often yields higher degree polynomial which is difficult to solve. Substitution will provide a better route.
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    Re: Interesting Equation! Solve for x.

    Actually there is an algebraic way of solving this problem. The solution is posted at Give me maths problems!. Generally, squaring or cubing to eliminate radical is not the a good method to solve equations containing radicals. Substitution will often solve the problem much more faster and easier.
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