If the minimum value of $\text{⨍ }\left(\mathrm{x}\right){\text{=2x}}^2\text{+αx+8}$ is the same as the maximum value of $\mathrm{g}\left(\mathrm{x}\right)=-3{\mathrm{x}}^2-4\mathrm{x}+{\mathrm{\alpha }}^2,$ then ${\mathrm{\alpha }}^2=$

$$\begin{gathered} {\text{⨍ (x)=2x}}^2\text{+αx+8} \\ {\text{⨍ }}^1\text{(x)=4x+α} \end{gathered}$$

$\text{⨍ (x)}$ is min or max

$$\begin{gathered} \Rightarrow {\text{⨍ }}^1\text{(x)=0} \\ \mathrm{x}=\frac{-\mathrm{\alpha }}{4} \end{gathered}$$

min or $\text{⨍ (x)}$ is

$$\begin{gathered} =2\left(\frac{{\mathrm{\alpha }}^2}{16}\right)+\mathrm{\alpha }\left(\frac{-\mathrm{\alpha }}{4}\right)+8 \\ =\frac{{\mathrm{\alpha }}^2-2{\mathrm{\alpha }}^2+64}{8} \\ =\frac{64-{\mathrm{\alpha }}^2}{8} \\ \mathrm{g}\left(\mathrm{x}\right)=-3{\mathrm{x}}^2-4\mathrm{x}+{\mathrm{\alpha }}^2 \\ {\mathrm{g}}^1\left(\mathrm{x}\right)=-64-4 \end{gathered}$$

g(x) is min or max

$$\begin{gathered} \Rightarrow {\mathrm{g}}^1\left(\mathrm{x}\right)=0 \Rightarrow \mathrm{x}=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ \mathrm{g}\left(\mathrm{x}\right)=-3\times \frac{4}{9}+\frac{8}{3}+{\mathrm{\alpha }}^2 \\ =\frac{4}{3}+{\mathrm{\alpha }}^2 \end{gathered}$$

given min $\text{⨍ (x)=}$ max g(x)

$$\begin{gathered} \frac{64-{\mathrm{\alpha }}^2}{8}=\frac{4}{3}+{\mathrm{\alpha }}^2 \\ 192-3{\mathrm{\alpha }}^2=32+24{\mathrm{\alpha }}^2 \\ \Rightarrow 27{\mathrm{\alpha }}^2=160\Rightarrow {\mathrm{\alpha }}^2=\frac{160}{27} \end{gathered}$$