<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] Your analysis is wrong becase $k$ depends on $x$: $k=\sqrt[4]x$. Instead, write $x=k^4$:
$$3k-2k^5-k^8+k^4=0$$
$$k^8+2k^5-k^4-3k=0$$
Descartes's rule of signs tells us there is exactly one positive root for this last polynomial. $k$ cannot be negative as it is the fourth root of $x$, so the lone root of the polynomial in $k$ translates to only one root for the original equation.
</s>