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.)
$\endgroup$ 32 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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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