The sum of the real roots of the equation x^4 | vedclass

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(C) Given the equation $x^4-2x^3+x-380=0$. By testing values,we find that $x=5$ is a root: $5^4-2(5^3)+5-380 = 625-250+5-380 = 380-380 = 0$. Next,we t

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(C) Given the equation $x^4-2x^3+x-380=0$. By testing values,we find that $x=5$ is a root: $5^4-2(5^3)+5-380 = 625-250+5-380 = 380-380 = 0$. Next,we test $x=-4$: $(-4)^4-2(-4)^3+(-4)-380 = 256+128-4-380 = 384-4-380 = 0$. Since $x=5$ and $x=-4$ are roots,we can divide the polynomial by $(x-5)(x+4) = x^2-x-20$. Performing polynomial division: $(x^4-2x^3+x-380) \div (x^2-x-20) = x^2-x+19$. The remaining roots are found by solving $x^2-x+19=0$. The discriminant $D = (-1)^2 - 4(1)(19) = 1 - 76 = -75$. Since $D Thus,the only real roots are $5$ and $-4$. The sum of the real roots is $5 + (-4) = 1$.

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

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