Question

Suppose $a,b$denote the distinct real roots of the quadratic polynomial $x^2+20x-2020=0$ and suppose $c,d$ denote the distinct complex roots of the quadratic polynomial $x^2-20x+2020$. Then the value of $ac\left(a-c\right)+ad\left(a-d\right)+bc\left(b-c\right)+bd\left(b-d\right)$ is

• A. a) 0
• B. b) 8000
• C. c) 8080
• D. d) 16000

Answer 1

The correct option is D

d) 16000

Correct option : d)

Explanation:

Step 1: Finding $a+b$and $ab$

Given $a,b$are the distinct real roots of the polynomial $x^2+20x-2020=0$.

Then sum of the roots $a+b=\frac{-20}{1}=-20$

The product of the roots $ab=\frac{-2020}{1}=-2020$

Step 2: Finding $c+d$and $cd$

Also given $c,d$are the distinct complex roots of the polynomial $x^2-20x+2020=0$

The sum of the roots $c+d=\frac{-\left(-20\right)}{1}=20$

The product of the roots $cd=\frac{2020}{1}=2020$

Step 3: Finding $ac\left(a-c\right)+ad\left(a-d\right)+bc\left(b-c\right)+bd\left(b-d\right)$

Now $ac\left(a-c\right)+ad\left(a-d\right)+bc\left(b-c\right)+bd\left(b-d\right)$

$=a^{2}c-a{c}^2+a^{2}d-a{d}^2+b^{2}c-b{c}^2+b^{2}d-b{d}^2$

$=a^2\left(c+d\right)+b^2\left(c+d\right)-c^2\left(a+b\right)-d^2\left(a+b\right)$

$=(a^2+b^2)\left(c+d\right)-\left(a+b\right)\left(c^2+d^2\right)$

$=\left({\left(a+b\right)}^2-2ab\right)\left(c+d\right)-\left({\left(c+d\right)}^2-2cd\right)\left(a+b\right)$

$=\left({\left(-20\right)}^2-2\left(-2020\right)\right)\left(20\right)-\left({\left(20\right)}^2-2\left(2020\right)\right)\left(-20\right)$

$=\left(400+4040\right)\left(20\right)-\left(400-4040\right)\left(-20\right)$

$=88800-72800$

$=16000$

Hence, $ac\left(a-c\right)+ad\left(a-d\right)+bc\left(b-c\right)+bd\left(b-d\right)=16000$