# Thread: how do you get roots of equations of fractional degree?

1. ## how do you get roots of equations of fractional degree?

Say you have equations like 3x^(7/3) + 2x^(2/5) - 6 = 0 , the degree of the equation is 7/3..
How do you solve for the roots? I know how to solve for the roots of higher degree equations like x^5, etc using rational root test but what about this? I ask because I encountered something like this in a calculus book. I made my calculator solve this and it gave me 1.1199, but I do not know if it is the only root or not.

2. ## Re: how do you get roots of equations of fractional degree?

Originally Posted by catenary
Say you have equations like 3x^(7/3) + 2x^(2/5) - 6 = 0 , the degree of the equation is 7/3..
How do you solve for the roots? I know how to solve for the roots of higher degree equations like x^5, etc using rational root test but what about this? I ask because I encountered something like this in a calculus book. I made my calculator solve this and it gave me 1.1199, but I do not know if it is the only root or not.
you can rewrite your expression as

$3x^{7/3} + 2x^{2/5} - 6=0$

$3x^{35/15} + 2x^{6/15}-6 =0$

$u=x^{1/15}$

$3u^{35} + 2u^6 - 6 = 0$

This can be solved numerically by your root finding algorithm of choice. In this case there will be 35 solutions....

The one real root is $u \approx 1.00758$

That solution $u_0$ can then be used to find $x$

$x=u^{15}$

$x=(1.00758)^{15} \approx 1.11991$

3. ## Re: how do you get roots of equations of fractional degree?

Thanks romsek. Your explanation is as clear as could possibly be. This is one less situation I have to worry about not knowing how to deal with.

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# how to solve equation with fractional degrees

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