Question

Two of the roots of the equation $x^3 - 9x^2 + 14x + 24 = 0$ are in the ratio $3 : 2$.
Then roots of the equation are

• A. 4, 6, -2
• B. 2, 3, -2
• C. 3, 2, -2
• D. 4, 6, -1

Answer 1

Answer: (D)

$x^{3} - 9x^{2} +14x +24 = 0 $
Let $\alpha,\beta,\gamma$ be the roots
given that two roots are in the ratio 3 : 2
Let $ \alpha:\beta = 3:2 \Rightarrow \alpha = 3k $ and $\beta =3k $
Sum of roots $= \alpha + \beta+\gamma=9$
$ \Rightarrow 3k + 2k + \gamma =9 \Rightarrow \gamma=9 -5k$
$ \alpha\beta+\beta\gamma+\gamma\alpha = 14$ and $ \alpha\beta\gamma = -24 $
$\Rightarrow 6k^{2} + 2k \left(9-5\right) + \left(9-5k\right) 3k = 14 $
$\Rightarrow 6k^{2} +18k - 10k^{2} +27 k -15k^{2} =14$
$ \Rightarrow - 19 k^{2} +45k = 14 \Rightarrow 19k^{2} -45k +14 = 0$
$ k = \frac{45\pm\sqrt{2025 - 4\times19\times14}}{38} $
$= \frac{45\pm\sqrt{961}}{38} = \frac{45\pm31}{38}$
$ k = \frac{76}{38} , \frac{14}{38} \Rightarrow k =2 , \frac{7}{19} $
If $k = 2 $ then $\alpha = 6 , \beta = 4 , \gamma = -1$
$ \therefore $ Roots of the given equation are 4, 6, -1