Suppose, a, b denote the distinct real roots of the quadratic polynomial x^2+20x−2020 and suppose c. d denote 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) 0 (B) 8000 (C) 8080 (D) 16000
Answers
Given : a and b are distinct real roots of x² + 20x - 2020 and c and d are distinct complex roots of x² - 20x + 2020.
To find : The value of ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)
solution : let's solve ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)
= a²c - ac² + a²d - ad² + b²c - bc² + b²d - bd²
= (a²c + a²d) + (b²c + b²d ) - (c²a + c²b) - (d²a + d²b)
= a²(c + d) + b²(c + d) - c²(a + b) - d²(a + b)
= (a² + b²)(c + d) - (c² + d²)(a + b)
= [(a + b)² - 2ab](c + d) - [(c + d)² - 2cd ] (a + b)
as a and b are roots of equation x² + 20x - 2020
so, a + b = -20 and ab = -2020
also c and d are roots of equation x² - 20x + 2020
so, c + d = 20 and cd = 2020
= [(-20)² - 2(-2020)] (20) - [(20)² - 2(2020)](-20)]
= [400 + 4040 ] 20 - [400 - 4040] (-20)
= 20[400 + 4040 + 400 - 4040]
= 20 × 800
= 16000
Therefore the option is (D) 16000
Answer:
(D)
Step-by-step explanation:
Given a quadratic equation
x²+20x -2020
a+b =-b'/a'=-20_____(1)
a*b = c'/a'=-2020 ____(2)
x²-20x + 2020
it's root are c and d
c+ d = 20 ____(3)
cd = 2020 _____(4)
=>ac(a-c) + ad ( a-d) + bc(b-c) + bd ( b -d)
=> a²c - ac² + a²d - ad² + b²c - bc² + b²d - bd²
=> a²c + a²d - ac² - ad² + b²c + b²d -bc²-bd²
=> a²(c+d) - a(c²-d²) + b²(c+d) - b(c²+d²)
=> (a² + b²)(c+d) -(a+b) {(c²+d²) }
=> we know that , (a+b)²-2ab = a²+b²
=> {(a+b)²-2ab} ( c+ d) -,{(a+b) (c+d)²-2cd}
=>{ (-20)² - 2(-2020) (20)}-(20){(20)²-2*2020}
=> 20 ( 400+4040 +400-4040)
=> 20× 800
= 16000 Answer .
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