JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Quadratic Equations and Inequalities

Suppose a, b denote the distinct real roots of the quadratic polynomial

x^{2} + 20 x - 2020

and suppose c, d denote the distinct complex roots of the quadratic polynomial

x^{2} - 20 x + 2020

. Then the value of

a c (a - c) + a d (a - d) + b c (b - c) + b d (b - d)

Answer: (D)

Solution

For the polynomial

x^{2} + 20 x - 2020 = 0

with roots

a, b \in R

, we have

a + b = - 20

and

ab = - 2020

.

For the polynomial

x^{2} - 20 x + 2020 = 0

with roots

c, d \in C

, we have

c + d = 20

and

c d = 2020

.

Expanding the given expression:

a c (a - c) + a d (a - d) + b c (b - c) + b d (b - d) = a^{2} (c + d) + b^{2} (c + d) - c^{2} (a + b) - d^{2} (a + b)

= (c + d) (a^{2} + b^{2}) - (a + b) (c^{2} + d^{2})

= (c + d) [(a + b)^{2} - 2 ab] - (a + b) [(c + d)^{2} - 2 c d]

= 20 [(- 20)^{2} + 4040] + (- 20) [(20)^{2} - 4040]

= 20 [400 + 4040] - 20 [400 - 4040]

= 20 \times 4440 - 20 \times (- 3640) = 20 (4440 + 3640) = 20 \times 8080

Wait, let me recalculate:

= 20 [400 + 4040] + 20 [400 - 4040] = 20 \times 800 = 16000

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Single correct option