Solve for x.

$\displaystyle x^3+1=2\sqrt[3]{2x-1} $

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- Jun 29th 2012, 09:27 AMthevinhInteresting Equation! Solve for x.
Solve for x.

$\displaystyle x^3+1=2\sqrt[3]{2x-1} $ - Jun 29th 2012, 11:08 AMHallsofIvyRe: 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. - Jun 29th 2012, 08:41 PMProve ItRe: Interesting Equation! Solve for x.
- Jun 29th 2012, 10:02 PMrichard1234Re: 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.

- Jun 29th 2012, 10:05 PMProve ItRe: Interesting Equation! Solve for x.
- Jul 1st 2012, 10:33 AMthevinhRe: Interesting Equation! Solve for x.
Actually there is an algebraic way to solve this problem without having to cube both sides. See solution posted at http://mathhelpforum.com/new-users/2...roblems-2.html. Squaring or cubing to eliminate radicals often yields higher degree polynomial which is difficult to solve. Substitution will provide a better route.

- Jul 1st 2012, 12:03 PMthevinhRe: Interesting Equation! Solve for x.
Actually there is an algebraic way of solving this problem. The solution is posted at http://mathhelpforum.com/new-users/2...roblems-2.html. 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.