Question

Answer this Brain Stromer :)​

Answer 1

Given Question :-

$\sf\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0 \: and \: \\ \sf \: c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0 \: then \:$

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =$

(a) 0

(b) 8000

(c) 8080

(d) 16000

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0}$

We know

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:ab = \dfrac{ - 2020}{1} = - 2020$

And

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:a + b = - \dfrac{20}{1} = - 20$

Also, given that

$\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0}$

$\rm \implies\:c + d = - \dfrac{( - 20)}{1} = 20$

and

$\rm \implies\:cd = \dfrac{2020}{1} = 2020$

Now, Consider

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)$

$\sf \: = {ca}^{2} - {ac}^{2} + {da}^{2} - {ad}^{2} + {cb}^{2} - {bc}^{2} + {db}^{2} - {bd}^{2}$

$\sf \: = {a}^{2}(c + d) + {b}^{2}(c + d) - {c}^{2}(a + b) - {d}^{2}(a + b)$

$\sf \: = (c + d)( {a}^{2} + {b}^{2}) - (a + b)( {c}^{2} + {d}^{2})$

$\sf \: = 20( {a}^{2} + {b}^{2}) + 20( {c}^{2} + {d}^{2})$

$\sf \: = 20\bigg[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\bigg]$

We know,

$\boxed{\tt{ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta \: }}$

So, using this, we get

$\sf \: = 20\bigg[ {(a + b)}^{2} - 2ab + {(c + d)}^{2} - 2cd\bigg]$

$\sf \: = 20\bigg[ {( - 20)}^{2} + 2(2020) + {(20)}^{2} - 2(2020)\bigg]$

$\sf \: = 20\bigg[ 400 + 400\bigg]$

$\sf \: = 20\bigg[ 800\bigg]$

$\sf \: = 16000$

Hence,

$\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}$

So, option (d) is correct.

Answer 2

option d.

hope it helps. ....