Interesting Equation! Solve for x.

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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Originally Posted by HallsofIvy

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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This is answered in full in another thread.

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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Originally Posted by richard1234

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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Actually there are only two more real solutions...

Quote:

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Originally Posted by richard1234

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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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.

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.