Q54 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
If $\alpha, \beta, \gamma$ are the roots of the equation $2 x^3-3 x^2+5 x-7$ $=0$, then $\sum \alpha^2 \beta^2=$
- Correct answerA$-\frac{17}{4}$
- B$\frac{17}{4}$
- C$-\frac{13}{4}$
- D$\frac{13}{4}$
Solution
$\alpha, \beta, \gamma$ are the roots of the equation
$\begin{aligned}
& 2 x^3-3 x^2+5 x-7=0 \\
& \Rightarrow \alpha+\beta+\gamma=\frac{3}{2} \Rightarrow \alpha \beta+\beta \gamma+\gamma \alpha=\frac{5}{2} \Rightarrow \alpha \beta \gamma=\frac{7}{2}
\end{aligned}$
By using identity $\begin{aligned} & \begin{aligned} &(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a \\ & \therefore(\alpha \beta+\beta \gamma+\gamma \alpha)^2=(\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2 \\ &+2 \alpha \beta^2 \gamma+2 \beta \gamma^2 \alpha+2 \gamma \alpha^2 \beta\end{aligned} \\ & \begin{aligned} & \Rightarrow(\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2=(\alpha \beta+\beta \gamma+\gamma \alpha)^2 \\ &-2 \alpha \beta \gamma(\alpha+\beta+\gamma)\end{aligned} \\ & \Rightarrow \sum \alpha^2 \beta^2=\left(\frac{5}{2}\right)^2-2 \times \frac{7}{2} \times \frac{3}{2}=\frac{-17}{4}\end{aligned}$
By using identity $\begin{aligned} & \begin{aligned} &(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a \\ & \therefore(\alpha \beta+\beta \gamma+\gamma \alpha)^2=(\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2 \\ &+2 \alpha \beta^2 \gamma+2 \beta \gamma^2 \alpha+2 \gamma \alpha^2 \beta\end{aligned} \\ & \begin{aligned} & \Rightarrow(\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2=(\alpha \beta+\beta \gamma+\gamma \alpha)^2 \\ &-2 \alpha \beta \gamma(\alpha+\beta+\gamma)\end{aligned} \\ & \Rightarrow \sum \alpha^2 \beta^2=\left(\frac{5}{2}\right)^2-2 \times \frac{7}{2} \times \frac{3}{2}=\frac{-17}{4}\end{aligned}$
Hence, the correct answer is option: 'A'.