Question

Suppose $a,b$ denotes the distinct real roots of the quadratic polynomial $x^2+20x-2020$ and suppose $c,d$ denotes 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

• A. $8080$
• B. $0$
• C. $8000$
• D. $16000$

Answer 1

The correct option is D $16000$

$x^2+20x-2020=0$ has two roots $a,b\in R.$
$a+b=-20$ & $a\cdot b=-2020$
& $x^2-20x+2020=0$ has two roots $c,d\in$ complex.
$c+d=20$ & $c\cdot d=2020$
now
$=ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$
$=a^{2}c-a{c}^2+a^{2}d-a{d}^2+b^{2}c-b{c}^2+b^{2}d-b{d}^2$
$=a^2(c+d)+b^2(c+d)-c^2(a+b)-d^2(a+b)$
$=(a^2+b^2)(c+d)-(a+b)(c^2+d^2)$
$=\left[\right(a+b{)}^2-2ab\left]\right(c+d)-[(c+d{)}^2-2cd](a+b)$
Put value $a,b,c$ & $d$ then,
$=\left[\right(-20{)}^2-2(-2020)\left]\right(20)-[(20{)}^2-2(2020\left)\right](-20)$
$=\left[\right(400+4040\left)\right]\left(20\right)-(-20)\left[\right(20{)}^2-4040]$
$=20[4440-3640]$
$=20\left[800\right]=16000$