5. Let planar vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ satisfy $|\boldsymbol{\alpha}+2 \boldsymbol{\beta}|=3,|2 \boldsymbol{\alpha}+3 \boldsymbol{\beta}|=4$, then the minimum value of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ is
Let $\left\{\begin{array}{l}u=\alpha+2 \boldsymbol{\beta}, \\ \boldsymbol{v}=2 \boldsymbol{\alpha}+3 \boldsymbol{\beta}\end{array} \Rightarrow\left\{\begin{array}{l}\alpha=2 \boldsymbol{v}-3 \boldsymbol{u} \\ \boldsymbol{\beta}=2 \boldsymbol{u}-\boldsymbol{v}\end{array} \Rightarrow \boldsymbol{\alpha} \cdot \boldsymbol{\beta}=(2 \boldsymbol{v}-3 \boldsymbol{u})(2 \boldsymbol{u}-\boldsymbol{v})\right.\right.$
$=7 \boldsymbol{u} \cdot \boldsymbol{v}-6 \boldsymbol{u}^{2}-2 \boldsymbol{v}^{2}=7 \boldsymbol{u} \cdot \boldsymbol{v}-86 \geqslant-170$, equality holds when $4 \boldsymbol{u}+3 \boldsymbol{v}=\mathbf{0}$
$\Rightarrow 10 \boldsymbol{\alpha}+17 \boldsymbol{\beta}=\mathbf{0}$. Therefore, the minimum value of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ is -170.