Question

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

Answer 1

The given polynomial can be expressed as:

$4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4(x - x_1)(x - x_2)(x - x_3)(x - x_4).$

Let $x_1 = 2i$ and $x_2 = -2i$. Substituting these values:

$64 - 64i + 68 - 24i + 9 = 4(2i - x_1)(2i - x_2)(2i - x_3)(2i - x_4).$

Simplify:

$141 - 88i \quad \dots \quad (1)$

Similarly, for $-2i$:

$64 + 64i + 68 + 24i + 9 = 4(-2i - x_1)(-2i - x_2)(-2i - x_3)(-2i - x_4).$

Simplify:

$141 + 88i \quad \dots \quad (2)$

Using the given condition:

$\frac{125}{16}m = \frac{141^2 + 88^2}{16}.$

Calculate:

$m = 221.$