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If $\alpha$ and $\beta$ are the roots of the equation $4 x^{2}+3 x+7=0$, then what is the value of $\left(\alpha^{-2}+\beta^{-2}\right)$ ?
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The correct answer is:
$-47 / 49$
Let $\alpha$ and $\beta$ be the roots of the equation. $4 x^{2}+3 x+7=0$
$\therefore \quad$ Sum $=\alpha+\beta=-\frac{3}{4}$ and Product $=\alpha \beta=\frac{7}{4}$
Consider, $\alpha^{2}+\beta^{-2}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}}$
$=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}=\frac{\frac{9}{16}-\frac{7}{2}}{\frac{49}{16}}$
$=\frac{\frac{9-56}{16}}{\frac{49}{16}}=\frac{-47}{16} \times \frac{16}{49}=-\frac{47}{49}$
$\therefore \quad$ Sum $=\alpha+\beta=-\frac{3}{4}$ and Product $=\alpha \beta=\frac{7}{4}$
Consider, $\alpha^{2}+\beta^{-2}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}}$
$=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}=\frac{\frac{9}{16}-\frac{7}{2}}{\frac{49}{16}}$
$=\frac{\frac{9-56}{16}}{\frac{49}{16}}=\frac{-47}{16} \times \frac{16}{49}=-\frac{47}{49}$
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