Find he number of real roots of the following equation. $$x^{\frac{1}{7}}-x^{\frac{1}{5}} = x^{\frac{1}{3}}-x^{\frac{1}{2}}.$$
$\bf{Attempt:}$ with Hit and Trial, Clearly $x=0,x=1$ are the roots of above equation.
Assuming $$f(x)=x^{\frac{1}{7}}-x^{\frac{1}{5}}-x^{\frac{1}{3}}+x^{\frac{1}{2}}$$
Then $$f'(x) = \frac{1}{7}x^{-\frac{6}{7}}-\frac{1}{5}x^{-\frac{4}{5}}-\frac{1}{3}x^{-\frac{2}{3}}+\frac{1}{2}x^{-\frac{1}{2}}.$$
Could some help me how to solve it, Thanks
There is another root.
Let $x=a^{210}$, where $a\geq0$.
Thus, we need to solve $$a^{30}-a^{42}=a^{70}-a^{105}$$ or $$a^{30}(a^{75}-a^{40}-a^{12}+1)=0$$ or $$a^{30}(a^{40}(a^{35}-1)-(a^{12}-1))=0$$ or $$a^{30}(a-1)(a^{40}(a^{34}+a^{33}+...+a+1)-(a^{11}+a^{10}+...+a+1))=0$$ and since $35\ne12$, by the Descartes's rule we see that there is another unique root on $(0,1)$.
About the Descartes's rule see here: https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs