Question

If $\alpha$ and $\beta$ are zeroes of $x^2-7x+10$ such that $\beta >\alpha$ find value of $\left(\frac{2a+3\beta }{\alpha \beta }\right)$

Answer 1

$\alpha ,\beta$ are roots of $k^2-7x+10=0$

$x^2-2x-5x+10=0$

$x(x-2)-5(x-2)=0$

$\Rightarrow (x-2)(x-5)=0$

$\Rightarrow x=2,5$

$\alpha =2,\beta =5$

$\frac{2\alpha +3\beta }{\alpha \beta }=\frac{2\left(2\right)+3\left(5\right)}{2\times 5}=\frac{19}{10}$.