If $\mathrm{\alpha }, \mathrm{\beta } ,1$ are roots of ${\mathrm{x}}^3 – 2{\mathrm{x}}^2 – 5\mathrm{x} + 6 = 0 (\mathrm{\alpha } > 1)$ then $3\mathrm{\alpha }+\mathrm{\beta }=$

$\begin{array}{l}\alpha ,\beta ,1arerootsof{x}^3-2x^2-5x+6=0\\ \therefore \alpha +\beta +1=2\Rightarrow \alpha +\beta =1--\left(1\right)\\ \alpha \beta \left(1\right)=-6\Rightarrow \alpha \beta =-6\\ \therefore \alpha -\beta =\sqrt{{(\alpha +\beta )}^2-4\alpha \beta }\Rightarrow \sqrt{1+24}\Rightarrow 5--\left(2\right)\\ \left(1\right)+\left(2\right)\Rightarrow 2\alpha =6\Rightarrow \alpha =3\\ \left(1\right)-\left(2\right)\Rightarrow 2\beta =-4\Rightarrow \beta =-2\\ \therefore 3\alpha +\beta =3\left(3\right)-2=7\end{array}$