Q52 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-a^2+b x-c$ $=0$ then, $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=$

  • A
    $\frac{b^2-3 a c}{c^2}$
  • B
    $\frac{b^2-a c}{c^2}$
  • Correct answer
    C
    $\frac{\mathrm{b}^2-2 \mathrm{ac}}{\mathrm{c}^2}$
  • D
    $\frac{b^2-4 a c}{e^2}$

Solution

Since $\alpha, \beta, \gamma$ are the roots of the equation $\begin{aligned} & x^3-a x^2+b x-c=0 \\ & \therefore \alpha+\beta+\gamma=\mathrm{a} \\ & \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{b} \\ & \alpha \beta \gamma=\mathrm{c}\end{aligned}$ Now, $\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{\alpha^2 \beta^2 \gamma^2}$ $\begin{aligned} & =\frac{(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma)}{(\alpha \beta \gamma)^2} \\ & =\frac{b^2-2 \times c \times a}{c^2}\end{aligned}$

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

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