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The product of real roots of the equation $4 x^4-24 x^3+$ $57 \mathrm{x}^2+18 \mathrm{x}-45=0$ if one of the root is $3+i \sqrt{6}$ is
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$-3 / 4$
Given that $3+i \sqrt{6}$ is one roots therefore $3-i \sqrt{6}$ is
also a root .Let $\alpha$ and $\beta$ are other two real roots
$\begin{aligned} & \therefore \text { product of roots }=\frac{-45}{4} \\ & \alpha \cdot \beta(3+\mathrm{i} \sqrt{6})(3-\mathrm{i} \sqrt{6})=\frac{-45}{4} \\ & \alpha \cdot \beta(9+6)=\frac{-45}{4} \\ & \alpha \cdot \beta=\frac{-3}{4}\end{aligned}$
also a root .Let $\alpha$ and $\beta$ are other two real roots
$\begin{aligned} & \therefore \text { product of roots }=\frac{-45}{4} \\ & \alpha \cdot \beta(3+\mathrm{i} \sqrt{6})(3-\mathrm{i} \sqrt{6})=\frac{-45}{4} \\ & \alpha \cdot \beta(9+6)=\frac{-45}{4} \\ & \alpha \cdot \beta=\frac{-3}{4}\end{aligned}$
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