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- Feb 14, 2012
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Determine all real \(\displaystyle x\) satisfying the equation \(\displaystyle \sqrt[5]{x^3+2x}=\sqrt[3]{x^5-2x}\)
Wolfram|Alpha also gives $x = \sqrt{2}$ as a real solution. One step closer to brute-forcing the problem!Just a note: Wolfram|Alpha will not give you correct solutions as written. It will give x = 0 as the only real solution. There are three real solutions. So no cheating!
-Dan
Actually that one ( and [tex]-\sqrt{2}[/tex]) aren't that hard to find. We know that there has to be more than x = 0, otherwise the problem is too simple. There are a number of 2's floating around there so it would seem wise to check solutions of the form [tex]2^{a/b}[/tex]. This yields the [tex]\pm \sqrt{2}[/tex] solutions, but really doesn't address the general problem.Wolfram|Alpha also gives $x = \sqrt{2}$ as a real solution. One step closer to brute-forcing the problem!![]()
I thought of that approach as well. The original equation is odd on both sides, so we know that any roots will be of the form y = x0, -x0. But note that when you raise the equation to the 15th power only the positive solutions survive.I find that when I raise both sides to the 15th power, divide through by $x^3$
Hey, it's been a long couple of days....Are you certin raising the original equation to the 15th power loses negative roots? Recall, we still found $x=-\sqrt{2}$ after doing so.