<s>[INST] Roots to an equation, $\frac{3-2x}{x-1}-\sqrt[4]{x^3}$ 
Find the number of roots to $\frac{3-2x}{x-1}-\sqrt[4]{x^3}=0$.

My method:
$\frac{3-2x}{x-1}-\sqrt[4]{x}$, we multiply $\sqrt[4]{x}$ and $x-1$ to both sides, getting $3\sqrt[4]{x}-2x\sqrt[4]{x}-x(x-1)=0$.  We let $\sqrt[4]{x}=k$.  Then we have $3k-2xk-x^2+x=0$.  Finding the discriminant ($b^2-4ac$), getting $4k^2+8k+1$, evidently greater than $0$, which means there are 2 roots.
Plotting this on Desmos gives 1 root only. How am I wrong?
 [/INST] Let $x=t^4$, where $t>0$.
Thus, $$t^7+2t^4-t^3-3=0$$ and we see that there is one changing of coefficients sing. 
Thus, the polynomial $t^7+2t^4-t^3-3$ has one positive root by https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs
In your reasoning you need to remember that $k\geq0,$ which gives that your method does not work.
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