Thread: 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.

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.

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.

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.

Actually there are only two more real solutions...

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.

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.

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.