"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 _________."

Question

Accepted Answer

"

First, observe that for a degree-4 polynomial

$P(x)=4x^4+8x^3-17x^2-12x+9$

with roots $x_1,\dots,x_4$, we have

$$\prod_{i=1}^4(x_i^2+4) =\prod_{i=1}^4\bigl[(x_i-2i)(x_i+2i)\bigr] =\frac{P(2i)\,P(-2i)}{4^2}\,. $$

Compute $P(2i)$:

$ P(2i) =4(2i)^4+8(2i)^3-17(2i)^2-12(2i)+9 =64-64i+68-24i+9 =141-88i. $

By conjugation,

$ P(-2i)=141+88i. $

Hence

$ \prod_{i=1}^4(x_i^2+4) =\frac{(141-88i)(141+88i)}{16} =\frac{141^2+88^2}{16} =\frac{19881+7744}{16} =\frac{27625}{16}. $

We are told

$ \prod_{i=1}^4(x_i^2+4)=\frac{125}{16}\,m =\frac{27625}{16}, $

so

$ 125\,m=27625 \quad\Longrightarrow\quad m=221. $

Answer: $ \displaystyle 221.$

"

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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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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