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(D) Given $x^2+20x-2020=0$ has roots $a, b$. By Vieta's formulas,$a+b = -20$ and $ab = -2020$.Given $x^2-20x+2020=0$ has roots $c, d$. By Vieta's form

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$16000$

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(D) Given $x^2+20x-2020=0$ has roots $a, b$. By Vieta's formulas,$a+b = -20$ and $ab = -2020$.Given $x^2-20x+2020=0$ has roots $c, d$. By Vieta's formulas,$c+d = 20$ and $cd = 2020$.We need to evaluate $E = ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$.Expanding the expression: $E = a^2c - ac^2 + a^2d - ad^2 + b^2c - bc^2 + b^2d - bd^2$.Grouping terms: $E = a^2(c+d) + b^2(c+d) - c^2(a+b) - d^2(a+b)$.$E = (c+d)(a^2+b^2) - (a+b)(c^2+d^2)$.Using $a^2+b^2 = (a+b)^2 - 2ab$ and $c^2+d^2 = (c+d)^2 - 2cd$:$a^2+b^2 = (-20)^2 - 2(-2020) = 400 + 4040 = 4440$.$c^2+d^2 = (20)^2 - 2(2020) = 400 - 4040 = -3640$.Substituting these values: $E = (20)(4440) - (-20)(-3640)$.$E = 88800 - 72800 = 16000$.

Suppose $a, b$ denote the distinct real roots of the quadratic polynomial $x^2+20x-2020$ and suppose $c, d$ denote the distinct complex roots of the quadratic polynomial $x^2-20x+2020$. Then the value of $ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$ is

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