Q56 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

If $\alpha, \beta, \gamma$ are the real roots of the equation $18 x^3-15 x^2-$ $4 x+4=0$ such that $\alpha=\beta$ and $\alpha>\gamma$, then $\alpha+\beta^2+\gamma^3=$

  • Correct answer
    A
    $\frac{71}{72}$
  • B
    $\frac{53}{54}$
  • C
    $\frac{89}{90}$
  • D
    $\frac{59}{60}$

Solution

$\begin{aligned} & \text 18 x^3-15 x^2-4 x+4=0 \\ & \alpha=\beta, \alpha>\gamma \\ & \alpha+\alpha+\gamma=\frac{15}{18}=\frac{5}{6} \\ & \Rightarrow \quad 2 \alpha+\gamma=\frac{5}{6}\end{aligned}$ $$ \begin{aligned} & \alpha^2+2 \alpha \gamma=\frac{-4}{18}=\frac{-2}{9} \\ & \alpha^2 \gamma=\frac{-4}{18}=\frac{-2}{9} \\ & \therefore \alpha^2+2 \alpha \gamma=\alpha^2 \gamma \\ & \Rightarrow \alpha(\alpha+2 \gamma)=\alpha^2 \gamma \Rightarrow \alpha+2 \gamma=\alpha \gamma \end{aligned} $$From (i), $\gamma=\frac{5}{6}-2 \alpha$ $$ \begin{aligned} & \Rightarrow \quad \alpha+2\left(\frac{5}{6}-2 \alpha\right)=\alpha\left(\frac{5}{6}-2 \alpha\right) \\ & \Rightarrow \quad \alpha+\frac{10}{6}-4 \alpha=\frac{5 \alpha}{6}-2 \alpha^2 \\ & \Rightarrow \quad 10-18 \alpha=5 \alpha-12 \alpha^2 \\ & \Rightarrow \quad 12 \alpha^2-23 \alpha+10=0 \\ & \Rightarrow \quad 12 \alpha^2-15 \alpha-8 \alpha+10=0 \\ & \Rightarrow \quad(3 \alpha-2)(4 \alpha-5)=0 \\ & \Rightarrow \quad \alpha=\frac{2}{3}, \frac{5}{4}, \gamma=\frac{-1}{2}, \frac{-5}{3} \end{aligned} $$ If $\alpha=\frac{2}{3} \Rightarrow \beta=\frac{2}{3}$ and $\gamma=\frac{-1}{2}$ $$ \alpha+\beta^2+\gamma^3=\frac{2}{3}+\frac{4}{9}-\frac{1}{8}=\frac{71}{72} . $$

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

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