The solution of the equation $2 x^3-x^2-22 x-24=0$, when two of the roots are in the ratio $3: 4$ is
The solution of the equation $2 x^3-x^2-22 x-24=0$, when two of the roots are in the ratio $3: 4$ is
- $3,4, \frac{1}{2}$
- $\frac{-3}{2},-2,4$
- $\frac{-1}{2}, \frac{3}{2}, 2$
- $\frac{-3}{2}, 2, \frac{5}{2}$
Solution
Given equation,
$
2 x^3-x^2-22 x-24=0
$
Put $x=-2$
$
\begin{aligned}
& 2(-2)^3-(-2)^2-22(-2)-24 \\
\Rightarrow \quad & -16-4+44-24=0
\end{aligned}
$
$\therefore x+2$ is root of equation $2 x^3-x^2-22 x-24=0$
$
\begin{aligned}
& 2 x^3-x^2-22 x-24=(x+2)\left(2 x^2-5 x-12\right) \\
& =(x+2)\left(2 x^2-8 x+3 x-12\right) \\
& =(x+2)(2 x+3)(x-4)
\end{aligned}
$
$\therefore$ Roots of equation are $-2, \frac{-3}{2}, 4$
Asked in: AP EAMCET 2021 (25 Aug Shift 1)
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