# 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.)

• 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 $f(\frac{1}{2})=\frac{1}{24} \left(8-7 \sqrt{2}\right)=-0.0791456$.
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 $x$.