JEE Challenger
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Value of Symmetric Polynomial Expression in Quadratic Roots

Assume a,ba, b represent the distinct real roots of the quadratic polynomial x2+20x−2020x^2 + 20x - 2020 and c,dc, d denote the distinct complex roots of the quadratic polynomial x2−20x+2020x^2 - 20x + 2020. Determine the value of ac(a−c)+ad(a−d)+bc(b−c)+bd(b−d)ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)

Options

A

00

B

80008000

C

80808080

D

1600016000

Correct

Step-by-Step Solution

To find the value of the given expression, let the expression be denoted by EE: E=ac(a−c)+ad(a−d)+bc(b−c)+bd(b−d)E = ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)

Step 1: Use Vieta's formulas for both quadratic polynomials

For the quadratic equation x2+20x−2020=0x^2 + 20x - 2020 = 0, the roots are aa and bb. By Vieta's formulas: a+b=−20a + b = -20 ab=−2020ab = -2020

For the quadratic equation x2−20x+2020=0x^2 - 20x + 2020 = 0, the roots are cc and dd. By Vieta's formulas: c+d=20c + d = 20 cd=2020cd = 2020

Step 2: Simplify the expression EE

Expand the given expression: E=a2c−ac2+a2d−ad2+b2c−bc2+b2d−bd2E = a^2c - ac^2 + a^2d - ad^2 + b^2c - bc^2 + b^2d - bd^2

Grouping the terms with a2,b2a^2, b^2 together and c2,d2c^2, d^2 together: E=a2(c+d)+b2(c+d)−c2(a+b)−d2(a+b)E = a^2(c + d) + b^2(c + d) - c^2(a + b) - d^2(a + b) E=(a2+b2)(c+d)−(c2+d2)(a+b)E = (a^2 + b^2)(c + d) - (c^2 + d^2)(a + b)

Step 3: Calculate a2+b2a^2 + b^2 and c2+d2c^2 + d^2

Using the algebraic identity x2+y2=(x+y)2−2xyx^2 + y^2 = (x + y)^2 - 2xy:

  1. For a2+b2a^2 + b^2: a2+b2=(a+b)2−2aba^2 + b^2 = (a + b)^2 - 2ab a2+b2=(−20)2−2(−2020)=400+4040=4440a^2 + b^2 = (-20)^2 - 2(-2020) = 400 + 4040 = 4440

  2. For c2+d2c^2 + d^2: c2+d2=(c+d)2−2cdc^2 + d^2 = (c + d)^2 - 2cd c2+d2=(20)2−2(2020)=400−4040=−3640c^2 + d^2 = (20)^2 - 2(2020) = 400 - 4040 = -3640

Step 4: Substitute the values into the expression EE

E=(4440)(20)−(−3640)(−20)E = (4440)(20) - (-3640)(-20) E=88800−72800=16000E = 88800 - 72800 = 16000

Thus, the value of the expression is 1600016000.

Correct Option: (D)