Solve $2x^3+3x^2-8x+3=0$, one root being double the another root.

Let root be $\alpha ,2\alpha ,\beta$

$\begin{array}{l}\alpha +2\alpha +\beta =-\frac{3}{2}\\ 3\alpha +\beta =-\frac{3}{2}\Rightarrow \beta \left(\alpha \right)=-\frac{8}{2}\\ 2{\alpha }^2+2\alpha (-\frac{3}{2}-3\alpha )+\alpha (-\frac{3}{2}-3\alpha )=-\frac{8}{2}\\ 2{\alpha }^2-3\alpha -6{\alpha }^2-\frac{3}{2}\alpha -3{\alpha }^2=-4\\ 4{\alpha }^2-6\alpha -12{\alpha }^2-3\alpha -6{\alpha }^2=-8\end{array}$

$\begin{array}{l}-14{\alpha }^2-9\alpha +8=-8\\ -14{\alpha }^2+9\alpha +8=D\\ 14{\alpha }^2+9\alpha -8=D\\ 14{\alpha }^2+16\alpha -7\alpha -8=D\end{array}$

$\begin{array}{l}2\alpha (7\alpha +8)-1(7\alpha +8)=D\\ (2\alpha -1)(7\alpha +8)=D\end{array}$

$\alpha =\frac{1}{2},\alpha =-\frac{8}{7}$$\text{case-}\left(\text{i}\right):\alpha =\frac{1}{2},\beta =-\frac{3}{2}-\frac{3}{2}=-3$

$\therefore$roots are $\frac{1}{2}1,-3$

$\text{case-}\left(\text{ii}\right)\alpha =-\frac{8}{7},\beta =-\frac{3}{2}+\frac{24}{7}=\frac{27}{14}$

Roots are $-\frac{8}{7},-\frac{16}{7},\frac{27}{14}$