We have,
$$
x^4-10 x^3+37 x^2-60 x+36=0
$$
Since, $\alpha, \alpha, \beta, \beta$ are the roots of the above equation.
$$
\begin{aligned}
& \therefore \quad \alpha+\alpha+\beta+\beta=\frac{-(-10)}{l}=10 \\
& \Rightarrow \quad \alpha+\beta=5 ...(i) \\
& \text { and } \quad(\alpha)(\alpha)(\beta)(\beta)=\frac{36}{1}...(ii) \\
& \Rightarrow \quad \alpha^2 \beta^2=36 \Rightarrow \alpha \beta=6 \\
&
\end{aligned}
$$
From Eqs. (i) and (ii), we have
$$
\begin{aligned}
\alpha(5-\alpha)=6 & \Rightarrow \alpha^2-5 \alpha+6=0 \\
\Rightarrow \quad \alpha=2,3 & \Rightarrow \beta=3,2
\end{aligned}
$$
Since $\alpha < \beta$, hence $\alpha=2, \beta=3$
$$
\begin{aligned}
\therefore \quad 2 \alpha+3 \beta & -2 \alpha \beta \\
& =2 \times 2+3 \times 3-2 \times 2 \times 3 \\
& =4+9-12=1
\end{aligned}
$$