<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] write your equation in the form
$$(3-2x)=(x-1)\sqrt[4]{x^3}(x-1)$$ for $x\ne 1$ and raise to the power four:
$$(3-2x)^4=x^3(x-1)^4$$
then you have to solve the equation
$$-{x}^{7}+4\,{x}^{6}-6\,{x}^{5}+20\,{x}^{4}-97\,{x}^{3}+216\,{x}^{2}-
216\,x+81
=0$$
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