• question_answer The two roots of an equation ${{x}^{3}}-9{{x}^{2}}+14x+24=0$ are in the ratio 3 : 2.  The roots will be [UPSEAT 1999] A) 6, 4, - 1 B) 6, 4, 1 C) - 6, 4, 1 D) - 6, - 4, 1

Let required roots are $3\alpha ,\,\,2\alpha ,\,\,\beta$ ($\because$ ratio of two roots are $3:2$) $\therefore \,\,\,\,\,\sum \alpha =3\alpha +2\alpha +\beta =\frac{-(-9)}{1}=9$ Þ $5\alpha +\beta =9$ ..?(i) $\sum \alpha \beta =3\alpha .2\alpha +2\alpha .\beta +\beta .3\alpha$$=14$ Þ $5\alpha \beta +6{{\alpha }^{2}}=14$ ?..(ii) and  $\sum \alpha \beta \gamma =3\alpha .2\alpha .\beta =-24$ Þ  $6{{\alpha }^{2}}\beta =-24$ or ${{\alpha }^{2}}\beta =-4$  ?..(iii) from (i),$\beta =9-5\alpha ,$put the value of $\beta$ in (ii) Þ $5\alpha (9-5\alpha )+6{{\alpha }^{2}}=14$   Þ  $19{{\alpha }^{2}}-45\alpha +14=0$ Þ $(\alpha -2)(19\alpha -7)=0$ $\therefore$ $\alpha =2$ or $\frac{7}{19}$ from (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. You will be redirected in 3 sec 