$\displaystyle
2x^3lnx+x^2-4x+a=0
$
i thought of building a taylor series around 0 for ln
but ln(0) is undefined
??
Let $\displaystyle f(x)=2x^3\ln x+x^2-4x+a, \ f0,\infty)\to\mathbf{R}$
We use the Rolle's sequence.
$\displaystyle f'(x)=6x^2\ln x+2x^2+2x-4$
$\displaystyle f'(x)=0\Rightarrow x=1$ and f' is increasing so it has an unique solution.
$\displaystyle \lim_{x\searrow 0}f(x)=a, \ f(1)=a-3, \ \lim_{x\to\infty}f(x)=\infty$
If $\displaystyle a\in(-\infty,0]$ then the Rolle's sequence is $\displaystyle - \ - \ +$ and f has a solution $\displaystyle x_1\in(1,\infty)$
If $\displaystyle a\in(0,3)$ then the Rolle's sequence is $\displaystyle + \ - \ +$ and f has two solutions $\displaystyle x_1\in(0,1), \ x_2\in(1,\infty)$
If $\displaystyle a=3$ then the Rolle's sequence is $\displaystyle + \ 0 \ +$ and f has the solution $\displaystyle x=1$
If $\displaystyle a>3$ then the Rolle's sequence is $\displaystyle + \ + \ +$ and f has no real solution.