I have something like
$a-ax+x^{0.8}-x^{0.2}=0$
with parameter a>0 and variable x>0.
I know by trial and error that the equation has three real roots for parameter a greater than certain value, otherwise it has only one root. x=1 is one obvious root. I want to find the value of parameter a that yields multiple roots. Is there a pen and paper solution to that, not using any computational algorithms like Newton's method etc.?
Let $y=x^{0.2}>0$, then the problem reduces to finding all $a>0$ such that the polynomial equation $a-ay^5+y^4-y=0$ has only 1 positive real root $y=1$.
For $y=1$ to be a double root, we must have $-5a(1)^4+4(1)^3-(1)=0$, so $a=\frac{3}{5}$. Putting this into the equation gives $$0=\frac{3}{5}-\frac{3}{5}y^5+y^4-y=-\frac{(y-1)^2}{5}(3y^3+y^2-y-3)=-\frac{(y-1)^3}{5}(3y^2+4y+3)$$
The quadratic $3y^2+4y+3$ has no real roots, so $a=\frac{3}{5}$ gives only 1 real root $y=1$ with multiplicity $3$.
Consider $y \not =1$, then we can write $$a=\frac{y^4-y}{y^5-1}=\frac{y^3+y^2+y}{y^4+y^3+y^2+y+1}$$
We know that $y=1$ is a triple root at $a=\frac{3}{5}$, so this motivates us to consider
\begin{align} & \frac{y^3+y^2+y}{y^4+y^3+y^2+y+1} \leq \frac{3}{5} \\ \Leftrightarrow & 3y^4+3y^3+3y^2+3y+3 \geq 5y^3+5y^2+5y \\ \Leftrightarrow & (y-1)^2(3y^2+4y+3) \geq 0 \end{align}
The function $f(y)=\frac{y^3+y^2+y}{y^4+y^3+y^2+y+1}$ achieves the values $\frac{3}{5}$ and $0$ at $y=1, 0$ respectively, so $f(y)$ can take all values from $0$ to $\frac{3}{5}$ exclusive (remember $y \not =0, 1$) since $f$ is continuous.
Thus $\forall a \in (0, \frac{3}{5})$, there exists $y>0, y \not =1$ s.t. $a=\frac{y^3+y^2+y}{y^4+y^3+y^2+y+1}$, and $\forall a \in [\frac{3}{5}, \infty)$, $y=1$ is the only positive real root.