First slide

Theory of equations

Question

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

Moderate

A

$6, 4, - 1$
B

$6, 4, 1$
C

$- 6, 4, 1$
D

$- 6, - 4, 1$

Solution

The given equation is $x^{3} - 9 x^{2} + 14 x + 24 = 0$
Let the required roots are $3 α, 2 α, β$         ( $∵$ ratio of two roots are 3:2)

$∴$ $\sum α = 3 α + 2 α + β = \frac{- (- 9)}{1} = 9$

$\Rightarrow$ $5 α + β = 9$                            …….(i)

   $\sum α β = 3 α .2 α + 2 α . β + β .3 α = 14$

$\Rightarrow$ $5 α β + 6 α^{2} = 14$                      ……(ii)

and    $α β γ = 3 α .2 α . β = - 24$

$\Rightarrow$ $6 a^{2} β = - 24$ or $α^{2} β = - 4$       .….(iii)

From Eq. (i), $β = 9 - 5 α$ , put the value of $β$ in Eq. (ii)

$\Rightarrow$ $5 α (9 - 5 α) + 6 α^{2} = 14$

$\Rightarrow$ $19 α^{2} - 45 α + 14 = 0$

$\Rightarrow$ $(α - 2) (19 α - 7) = 0$

$∴$ $α = 2$ or $\frac{7}{19}$

$∴$ from Eq. (i), if $α = 2$ , then $β = 9 - 5 \times 2 = - 1$

$∵ α = 2, β = - 1$ satisfy the equation (iii) so required roots are $6, 4, - 1$ .

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