Q54 / 160 · AP EAMCET 2022 (08 Jul Shift 1) · Mathematics

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

  • A
    $-5 / 16$
  • B
    $5 / 16$
  • C
    $3 / 4$
  • Correct answer
    D
    $-3 / 4$

Solution

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 $\begin{aligned} & \therefore \text { product of roots }=\frac{-45}{4} \\ & \alpha \cdot \beta(3+\mathrm{i} \sqrt{6})(3-\mathrm{i} \sqrt{6})=\frac{-45}{4} \\ & \alpha \cdot \beta(9+6)=\frac{-45}{4} \\ & \alpha \cdot \beta=\frac{-3}{4}\end{aligned}$

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

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