Question

Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 & c, d are roots of x² - 20x + 2022 then the value of ac(a - c) ad(a - d) + bc(b - c) + (b - d)

Choose the correct option
(a) 0
(b) 8000
(c) 8080
(d) 16000​

Answer 1

Correct 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=(Constant)/(coefficient\ of\ x^(2))}}`

`\rm \implies\:ab = ( - 2020)/(1) = - 2020`

And

`\boxed{\red{\sf Sum\ of\ the\ zeroes=(-coefficient\ of\ x)/(coefficient\ of\ x^(2))}}`

`\rm \implies\:a + b = - (20)/(1) = - 20`

Also, given that

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

`\rm \implies\:c + d = - (( - 20))/(1) = 20`

and

`\rm \implies\:cd = (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.