If $\alpha ,\beta ,\gamma ,\delta$  are the roots of $x^4-2x^3-8x^2+18x-9=0$  such that $\alpha +\beta =0$  then $3\gamma +4\delta =$

The given equation is $x^4-2x^3-8x^2+18x-9=0$

$\alpha ,\beta ,\gamma ,\delta \text{are the roots of the equation}$

$\text{Given that }\alpha +\beta =0$

$s_1:\alpha +\beta +\gamma +\delta =-\left(-2\right)$ $\Rightarrow \gamma +\delta =2,$ ….. ( 1 )

$s_2=\left(\alpha +\beta \right)\left(\gamma +\delta \right)+\alpha \beta +\gamma \delta =-8$ $\Rightarrow \alpha \beta +\gamma \delta =-8$

$s_3=\left(\alpha +\beta \right)\gamma \delta +\left(\gamma +\delta \right)\alpha \beta =-18$ $\Rightarrow \alpha \beta =-9$

$s_4=\alpha \beta \gamma \delta =9$  $\Rightarrow \gamma \delta =1$ ….. ( 2 )

From ( 1 ) & ( 2 ), we get

$\gamma =\frac{1}{\delta }$

$\Rightarrow \frac{1}{\delta }+\delta =2\Rightarrow \delta =1$

$\therefore \gamma =1,\delta =1$