The sum of the complex roots of the equation | vedclass

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(D) Given equation: $x^4-2x^3+x-380=0$. By checking the roots,for $x=5$: $5^4-2(5^3)+5-380 = 625-250+5-380 = 0$. So,$x=5$ is a root. For $x=-4$: $(-4)

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(D) Given equation: $x^4-2x^3+x-380=0$. By checking the roots,for $x=5$: $5^4-2(5^3)+5-380 = 625-250+5-380 = 0$. So,$x=5$ is a root. For $x=-4$: $(-4)^4-2(-4)^3+(-4)-380 = 256+128-4-380 = 0$. So,$x=-4$ is a root. Let the four roots be $x_1, x_2, x_3, x_4$. We have $x_1=5$ and $x_2=-4$. Let $x_3$ and $x_4$ be the complex roots. From the relation between roots and coefficients,the sum of all roots is given by $-\frac{	ext{coefficient of } x^3}{	ext{coefficient of } x^4} = -\frac{-2}{1} = 2$. Thus,$x_1+x_2+x_3+x_4 = 2$. Substituting the known roots: $5-4+x_3+x_4 = 2$. $1+x_3+x_4 = 2$. Therefore,$x_3+x_4 = 1$.

The sum of the complex roots of the equation $x^4-2x^3+x-380=0$ is

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