Question

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

• A. -6, 4, 1
• B. 6, 4, 1
• C. 6, 4, -1
• D. -6, -4, 1

Answer 1

The correct option is C

6, 4, -1

Let required roots are $3\alpha ,2\alpha ,\beta$

($\because$ ratio of two roots are 3:2)

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

$\Rightarrow 5\alpha +\beta =9\cdots \cdots \left(i\right)\sum \alpha \beta =3\alpha .2\alpha +2\alpha .\beta +\beta .3\alpha =14\Rightarrow 5\alpha \beta +6{\alpha }^2=14and\sum \alpha \beta \gamma =3\alpha .2\alpha .\beta =-24$

$\Rightarrow 6{\alpha }^2\beta =-24or{\alpha }^2\beta =-4From\left(i\right),\beta =9-5\alpha ,putthevalueof\beta in\left(ii\right)\Rightarrow 5\alpha (9-5\alpha )+6{\alpha }^2=14\Rightarrow 19{\alpha }^2-45\alpha +14=0\Rightarrow (\alpha -2)(19\alpha -7)=0$

$\therefore \alpha =2or\frac{7}{19}\therefore from\left(i\right),if\alpha =2,then\beta =9-5\times 2=-1$

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