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

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4-2.Quadratic Equations and Inequations

hard

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

$6, 4, -1$

$6, 4, 1$

$-6, 4, 1$

$-6, -4, 1$

Solution

(a)Let required roots are $3\alpha ,\,\,2\alpha ,\,\,\beta $
( ratio of two roots are $3:2$)
$\therefore \,\,\,\,\,\sum \alpha = 3\alpha + 2\alpha + \beta = \frac{{ – ( – 9)}}{1} = 9$
==> $5\alpha + \beta = 9$ ..…$(i)$
$\sum \alpha \beta = 3\alpha .2\alpha + 2\alpha .\beta + \beta .3\alpha $$ = 14$
==> $5\alpha \beta + 6{\alpha ^2} = 14$ …..$(ii)$
and $\sum \alpha \beta \gamma = 3\alpha .2\alpha .\beta = – 24$
==> $6{\alpha ^2}\beta = – 24$ or ${\alpha ^2}\beta = – 4$ …..$(iii)$
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 $\frac{7}{{19}}$

$\therefore $ from $(i)$ , if $\alpha = 2,$ then$\beta = 9 – 5 \times 2$ $= -1$

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

Standard 11

Mathematics

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Solution

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