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.You need to login to perform this action.
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