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 $\alpha \beta \gamma =-24$
$\Rightarrow 6k^2+2k\left(9-5\right)+\left(9-5k\right)3k=14$
$\Rightarrow 6k^2+18k-10k^2+27k-15k^2=14$
$\Rightarrow -19k^2+45k=14\Rightarrow 19k^2-45k+14=0$
$k=\frac{45\pm \sqrt{2025-4\times 19\times 14}}{38}$
$=\frac{45\pm \sqrt{961}}{38}=\frac{45\pm 31}{38}$
$k=\frac{76}{38},\frac{14}{38}\Rightarrow k=2,\frac{7}{19}$
If $k=2$ then $\alpha =6,\beta =4,\gamma =-1$
$\therefore$ Roots of the given equation are 4, 6, -1