Let x_1, x_2, x_3, x_4 be the roots of the eq | vedclass

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(D) Let $P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$.We want to evaluate the product $S = (4+x_1^2)(4+x_2^2)(4+x_3^2)(4+x_4^

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$221$

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(D) Let $P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$.We want to evaluate the product $S = (4+x_1^2)(4+x_2^2)(4+x_3^2)(4+x_4^2)$.Note that $4+x_k^2 = (2i+x_k)(-2i+x_k) = (x_k - 2i)(x_k + 2i)$.Thus,$S = \prod_{k=1}^4 (x_k - 2i) \prod_{k=1}^4 (x_k + 2i) = \prod_{k=1}^4 (2i - x_k) \prod_{k=1}^4 (-2i - x_k)$.From $P(x) = 4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$,we have $\prod_{k=1}^4 (x-x_k) = \frac{P(x)}{4}$.So,$\prod_{k=1}^4 (2i - x_k) = \frac{P(2i)}{4}$ and $\prod_{k=1}^4 (-2i - x_k) = \frac{P(-2i)}{4}$.$P(2i) = 4(2i)^4 + 8(2i)^3 - 17(2i)^2 - 12(2i) + 9 = 4(16) + 8(-8i) - 17(-4) - 24i + 9 = 64 - 64i + 68 - 24i + 9 = 141 - 88i$.$P(-2i) = 4(-2i)^4 + 8(-2i)^3 - 17(-2i)^2 - 12(-2i) + 9 = 4(16) + 8(8i) - 17(-4) + 24i + 9 = 64 + 64i + 68 + 24i + 9 = 141 + 88i$.$S = \frac{P(2i)}{4} 	imes \frac{P(-2i)}{4} = \frac{(141-88i)(141+88i)}{16} = \frac{141^2 + 88^2}{16} = \frac{19881 + 7744}{16} = \frac{27625}{16}$.Given $S = \frac{125}{16}m$,we have $\frac{27625}{16} = \frac{125}{16}m$.$m = \frac{27625}{125} = 221$.

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

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