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

Correct Answer: A

Solution :

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.