# Solving Equation with Fractional Exponents

In the final solution to calculus problem I'm struggling to solve the following equation for x:

$(1-x)^{5/2}-(5/3)(1-x)^{3/2}=-1/3$

Further, while I hope I'm just having a "brain glitch", if anyone can suggest where I might be weak algebraically in failing to solve this on my own, I would appreciate any advice/suggestions.

Thanks.

(Note: I already know the answer via my wonderful calculator, but I do not like relying on it unnecessarily.)

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For this type of problem, you won't get rational values of $x$. Let $u = (1 - x)^{1/2}$. Then, we have $u^5 - 5/3 \cdot u^3 = -1/3$, which is $3u^5 - 5u^3 + 1 = 0$. Then, $u$ is irrational because there are no rational roots that work. Try to use Rational Root Theorem, and you see that no rational roots exist. – NasuSama Mar 8 '14 at 1:40
The equation can be solved numerically. It is possible that there was a glitch in the calculation that led to this equation, for it is common when making up probems to set up numbers so that roots are "nice." – André Nicolas Mar 8 '14 at 1:44
Thank you all for the assistance! This problem is presumably one of the more challenging, higher-numbered problems in the exercise set I was attempting covering aspects of integration. I am relieved to know that it is genuinely a tricky problem and that I wasn't just missing something obvious. After all, that final equation at a glance appears much simpler than the initial problem! – juanproya Mar 8 '14 at 16:23

By inspection, you could notice that, if $$f(x)=(1-x)^{5/2}-(5/3)(1-x)^{3/2}+1/3$$ $f(-1)=\frac{1}{3}+\frac{2 \sqrt{2}}{3}$ which is positive, $f(0)=-\frac{1}{3}$ which is negative and $f(1)=\frac{1}{3}$ which is positive. Then, the equation shows at least two roots, one between $-1$ and $0$, and another one between $0$ and $+1$. You could even go further and notice that the roots are close to $-\frac{1}{2}$ and $\frac{1}{2}$ since $f(-\frac{1}{2})=\frac{1}{24} \left(8-3 \sqrt{6}\right)=0.0271471$ and $f(\frac{1}{2})=\frac{1}{24} \left(8-7 \sqrt{2}\right)=-0.0791456$.

As mentioned in other answers and comments, getting accurate solutions will require purely numerical methods (these are quite simple is you know derivatives).

Let me know if you want me to elaborate in this direction.

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I'd suggest writing $y = (1-x)^{1/2}$. Then you get $3y^5 -5y^3 +3 = 0$. Still not easy, but a vast improvement. You can solve for $y$, but I think you'll have to do this with numerical methods. Then you can get $x$.

This seems much too hard for a calculus class. I'd guess that the question has a typo.

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