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- Feb 14, 2012

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- Aug 30, 2012

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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

-Dan

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- Jan 26, 2012

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Wolfram|Alpha also gives $x = \sqrt{2}$ as a real solution. One step closer to brute-forcing the problem!Just a note: Wolfram|Alpha willnotgive you correct solutions as written. It will give x = 0 as the only real solution. There are three real solutions. So no cheating!

-Dan

- Aug 30, 2012

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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!

Then we have to prove that there are only three real solutions...

-Dan

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\(\displaystyle t^{11}-10t^9+40t^7-80t^5+79t^3-6t^2-44t-8=0\)

and the rational roots theorem then allows me to factor this as:

\(\displaystyle (t+1)(t-2)(t^9+t^8-7t^7-5t^6+21t^5+11t^4-27t^3-5t^2+20t+4)=0\)

So now we have the real roots \(\displaystyle x=\pm\sqrt{2}\), and I am left to show that:

\(\displaystyle t^9+t^8-7t^7-5t^6+21t^5+11t^4-27t^3-5t^2+20t+4=0\)

has no positive real roots.

Descartes' rule of signs tells us there are 0, 2, or 4 positive real roots, so no help there.

- Aug 30, 2012

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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 $x^3$

eg

[tex](x^2 + 2)^3 = x^2(x^4 - 2)^5[/tex]

Both sides must be positive so the RHS implies that [tex]x > 2^{1/4}[/tex]

Nice thought about that t substitution...I left it in terms of x^2 but for some reason didn't see that.

-Dan

Edit: My graph shows that there is a real solution for t (as there must be), but it's negative. So x is imaginary for this case.

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Are you certin raising the original equation to the 15th power loses negative roots? Recall, we still found $x=-\sqrt{2}$ after doing so.

- Aug 30, 2012

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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.

Thanks for the catch!

-Dan

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- Feb 14, 2012

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Now without any further ado, here is my solution to this problem...

From the given equation, we raise both sides to the 15th power and get:

\(\displaystyle \sqrt[5]{x^3+2x}=\sqrt[3]{x^5-2x}\)

\(\displaystyle (x^3+2x)^3=(x^5-2x)^5\)

We then factor out the common factor from both sides of the equation and simplify:

\(\displaystyle (x^2+2)^3=x^2(x^4-2)^5\)

If we let \(\displaystyle k=x^2\), the above equation becomes \(\displaystyle (k+2)^3=k(k^2-2)^5\).

My first instinct when I saw this form of equation was to relate the LHS of the equation as the product of two factors, i.e. \(\displaystyle a=b(c)\) and hoped that if \(\displaystyle a=b\) then \(\displaystyle c=1\)...thus, we try to algebraically modify the RHS of the equation and the best plan is to transfrom the RHS of the equation be the product of something raised to the third power and some other factor. Let's see...

\(\displaystyle (k+2)^3=k(k^2-2)^5\)

\(\displaystyle (k+2)^3=\frac{k^2}{k^2}\left(k(k^2-2)^3(k^2-2)^2\right)\)

\(\displaystyle (k+2)^3=k^3(k^2-2)^3\left(\frac{k^2-2}{k}\right)^2\)

\(\displaystyle (k+2)^3=(k(k^2-2))^3\left(\frac{k^2-2}{k}\right)^2\)

Now, let \(\displaystyle k+2=k(k^2-2)\) and we should expect to find such a solution where when \(\displaystyle k+2=k(k^2-2)\) is true for some k values, and when we substitute those k value(s) to other expression, we will end up with \(\displaystyle \left(\frac{k^2-2}{k}\right)^2=1\).

We solve the equation \(\displaystyle k+2=k(k^2-2)\) by trial and error method and obtain:

\(\displaystyle k^3-3k-2=0\)

\(\displaystyle (k+1)^2(k-2)=0\)

Thus, \(\displaystyle k=-1\) or \(\displaystyle k=2\) but \(\displaystyle k=x^2(\ge0)\), we then eliminate \(\displaystyle k=-1\) as one of the solutions to this problem.

In order to tell if \(\displaystyle k=2\) is the solution to the equation \(\displaystyle (k+2)^3=(k(k^2-2))^3\left(\frac{k^2-2}{k}\right)^2\), we still need to substitute \(\displaystyle k=2\) into another factor of the RHS's expression and make sure we gotten 1 from it.

So, when \(\displaystyle k=2\), we see that \(\displaystyle \left(\frac{k^2-2}{k}\right)^2=\left(\frac{2^2-2}{2}\right)^2=(1)^2=1\) which proves that \(\displaystyle k=x^2=2\) or \(\displaystyle x=\pm \sqrt{2}\) are the solutions to the original radical equation.

Also, observe that x=0 is another solution to the equation

\(\displaystyle \sqrt[5]{x^3+2x}=\sqrt[3]{x^5-2x}\) as well.

By combining all the x values that we found, the solutions to this problem are \(\displaystyle x=0\) and \(\displaystyle x=\pm \sqrt{2}\).