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?
On
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 $$-{x}^{7}+4\,{x}^{6}-6\,{x}^{5}+20\,{x}^{4}-97\,{x}^{3}+216\,{x}^{2}- 216\,x+81 =0$$
On
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.
On
Since $k$ is a function of $x$, the coefficients of your quadratric equation are functions of $x$.
The quadratic formula is useless in that context, since that would be solving for $x$ in terms of $x$.
Here's how I would approach the problem . . .
Let $f(x)={\displaystyle{{\small{\frac{3-2x}{x-1}}}-\sqrt[4]{x^3}}}$.
You want to count the number of real zeros of $f$.
Rather than trying to solve the equation $f(x) = 0$, instead, analyze the behavior of $f$.
The domain of $f$ is $[0,1)\cup(1,\infty)$.
On the interval $[0,1)$, we have $$\frac{3-2x}{x-1}<0$$ $$-\sqrt[4]{x^3} \le 0$$ hence $f(x) < 0$.
Thus, $f$ has no real zeros in the interval $[0,1)$.
On the interval $(1,\infty)$, the functions $$\frac{3-2x}{x-1}$$ $$-\sqrt[4]{x^3}$$ are both strictly decreasing, hence $f$ is strictly decreasing.
It follows that $f$ has at most one real root.
Letting $x$ approach $1$ from the right, we get \begin{align*} \lim_{x\to 1^{+}} f(x) &=\;\lim_{x\to 1^{+}} \frac{3-2x}{x-1}-\sqrt[4]{x^3}\\[4pt] &=\;\lim_{x\to 1^{+}} \frac{3-2x}{x-1}-\lim_{x\to 1^{+}}\sqrt[4]{x^3}\\[4pt] &=\;\infty-1\\[4pt] &=\;\infty\\[4pt] \end{align*} Letting $x$ approach $\infty$, we get \begin{align*} \lim_{x\to \infty} f(x) &=\;\lim_{x\to \infty} \frac{3-2x}{x-1}-\sqrt[4]{x^3}\\[4pt] &=\;\lim_{x\to \infty} \frac{3-2x}{x-1}-\lim_{x\to \infty}\sqrt[4]{x^3}\\[4pt] &=\;-2-\infty\\[4pt] &=\;-\infty\\[4pt] \end{align*} Since $f$ is continuous on $(1,\infty)$, it follows that $f$ has at least one real root.
Therefore $f$ has exactly one real root.
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.