Q54 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

$\alpha, \beta, \gamma$ are the roots of the equation $x^3+3 x^2-10 x-24=0$. If $\alpha\gt\beta\gt\gamma$ and $\alpha^3+3 \beta^2-10 \gamma-24=11 k$, then $k=$

  • A
    1
  • B
    11
  • Correct answer
    C
    5
  • D
    55

Solution

$x^3+3 x^2-10 x-24=0, \alpha, \beta, \gamma$ are its roots $x=3$ is one of its root (by hit and trial) On dividing by $(x-3)$, we get $(x-3)\left(x^2+6 x+8\right)=0$ Other roots are factors of $x^2+6 x+8=(x+4)(x+2)$ i.e. $x=-4, x=-2 \Rightarrow \alpha=3, \beta=-2, \gamma=-4(\alpha\gt\beta\gt\gamma)$ $\Rightarrow 55=11 K \Rightarrow K=5$.

Hence, the correct answer is option: 'C'.

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