Let $x_1, x_2, x_3, x_4$ be the solution of the equation $4 x^4+8 x^3-17 x^2-12 x+9=0$ and $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$. Then the value of $m$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 4 x^4+8 x^3-17 x^2-12 x+9 \\ & =4\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right) \end{aligned}$
Put $x=2 i \&-2 i$ $\begin{aligned} & 64-64 i+68-24 i+9=\left(2 i-x_1\right)\left(2 i-x_2\right)\left(2 i-x_3\right) \\ & \left(2 i-x_4\right) \\ & =141-88 i ....(1)\\ & 64+64 i+68+24 i+9=4\left(-2 i-x_1\right)\left(-2 i-x_2\right)(-2 i \\ & \left.-x_3\right)\left(-2 i-x_4\right) \\ & =141+88 i....(2) \end{aligned}$ $\begin{aligned} & \frac{125}{16} \mathrm{~m}=\frac{141^2+88^2}{16} \\ & \mathrm{~m}=221 \end{aligned}$

Let $x_1, x_2, x_3, x_4$ be the solution of the equation $4 x^4+8 x^3-17 x^2-12 x+9=0$ and…

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