The two roots of an equation $x^3-9x^2+14x+24=0$  are in the ratio 3:2, The roots will be

The given equation is $x^3-9x^2+14x+24=0$

Let the  required roots are $3\alpha ,2\alpha ,\beta$         ($\because$  ratio of two roots are 3:2)

$\therefore$ ${\displaystyle \sum \alpha }=3\alpha +2\alpha +\beta =\frac{-\left(-9\right)}{1}=9$

$\Rightarrow$ $5\alpha +\beta =9$                            …….(i)

   ${\displaystyle \sum \alpha \beta =3\alpha .2\alpha +2\alpha .\beta +\beta .3\alpha =14}$

$\Rightarrow$ $5\alpha \beta +6{\alpha }^2=14$                      ……(ii)

and    $\alpha \beta \gamma =3\alpha .2\alpha .\beta =-24$

$\Rightarrow$ $6a^2\beta =-24$  or ${\alpha }^2\beta =-4$       .….(iii)

From Eq. (i), $\beta =9-5\alpha$ , put the value of $\beta$  in Eq. (ii)

$\Rightarrow$ $5\alpha \left(9-5\alpha \right)+6{\alpha }^2=14$

$\Rightarrow$ $19{\alpha }^2-45\alpha +14=0$

$\Rightarrow$ $\left(\alpha -2\right)\left(19\alpha -7\right)=0$

$\therefore$  $\alpha =2$  or $\frac{7}{19}$

$\therefore$  from Eq. (i), if $\alpha =2$ , then $\beta =9-5\times 2=-1$

$\because \alpha =2,\beta =-1$  satisfy the equation (iii) so required roots are $6,4,-1$ .