The two roots of an equation {x^3} - 9{x^2} + | vedclass

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Question

(a)Let required roots are $3\alpha ,\,\,2\alpha ,\,\,\beta $
( ratio of two roots are $3:2$)
$	herefore \,\,\,\,\,\sum \alpha = 3\alpha + 2\alpha +

Vedclass · Accepted Answer

$6, 4, -1$

Vedclass · Answer

(a)Let required roots are $3\alpha ,\,\,2\alpha ,\,\,\beta $
( ratio of two roots are $3:2$)
$	herefore \,\,\,\,\,\sum \alpha = 3\alpha + 2\alpha + \beta = \frac{{ - ( - 9)}}{1} = 9$
==> $5\alpha + \beta = 9$ ..&hellip;$(i)$
$\sum \alpha \beta = 3\alpha .2\alpha + 2\alpha .\beta + \beta .3\alpha $$ = 14$
==> $5\alpha \beta + 6{\alpha ^2} = 14$ &hellip;..$(ii)$
and $\sum \alpha \beta \gamma = 3\alpha .2\alpha .\beta = - 24$
==> $6{\alpha ^2}\beta = - 24$ or ${\alpha ^2}\beta = - 4$ &hellip;..$(iii)$
from $(i)$,$\beta = 9 - 5\alpha ,$put the value of $\beta $ in $(ii)$
==> $5\alpha (9 - 5\alpha ) + 6{\alpha ^2} = 14$
==>$19{\alpha ^2} - 45\alpha  + 14 = 0$

==>$(\alpha  - 2)(19\alpha  - 7) = 0$

$\alpha  = 2$ or $\frac{7}{{19}}$

$	herefore $ from $(i)$ ,  if $\alpha  = 2,$ then$\beta  = 9 - 5 	imes 2$ $= -1$

$\alpha  = 2,\beta  =  - 1$  satisfy the equation $(iii)$ so required  roots are $6, 4, -1.$

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

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