Correct Question :-
[tex] \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 \: [/tex]
[tex] \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = [/tex]
(a) 0
(b) 8000
(c) 8080
(d) 16000
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0}[/tex]
We know
[tex]\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}[/tex]
[tex]\rm \implies\:ab = \dfrac{ - 2020}{1} = - 2020[/tex]
And
[tex]\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}[/tex]
[tex]\rm \implies\:a + b = - \dfrac{20}{1} = - 20[/tex]
Also, given that
[tex]\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0}[/tex]
[tex]\rm \implies\:c + d = - \dfrac{( - 20)}{1} = 20[/tex]
and
[tex]\rm \implies\:cd = \dfrac{2020}{1} = 2020[/tex]
Now, Consider
[tex] \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)[/tex]
[tex] \sf \: = {ca}^{2} - {ac}^{2} + {da}^{2} - {ad}^{2} + {cb}^{2} - {bc}^{2} + {db}^{2} - {bd}^{2} [/tex]
[tex] \sf \: = {a}^{2}(c + d) + {b}^{2}(c + d) - {c}^{2}(a + b) - {d}^{2}(a + b)[/tex]
[tex] \sf \: = (c + d)( {a}^{2} + {b}^{2}) - (a + b)( {c}^{2} + {d}^{2})[/tex]
[tex] \sf \: = 20( {a}^{2} + {b}^{2}) + 20( {c}^{2} + {d}^{2})[/tex]
[tex] \sf \: = 20\bigg[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\bigg][/tex]
We know,
[tex]\boxed{\tt{ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta \: }}[/tex]
So, using this, we get
[tex] \sf \: = 20\bigg[ {(a + b)}^{2} - 2ab + {(c + d)}^{2} - 2cd\bigg][/tex]
[tex] \sf \: = 20\bigg[ {( - 20)}^{2} + 2(2020) + {(20)}^{2} - 2(2020)\bigg][/tex]
[tex] \sf \: = 20\bigg[ 400 + 400\bigg][/tex]
[tex] \sf \: = 20\bigg[ 800\bigg][/tex]
[tex] \sf \: = 16000[/tex]
Hence,
[tex]\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}[/tex]
So, option (d) is correct.
