Question

Prove that (a²+b²)(c²+d²) = (ac+bd)² + (a-bc)².
(Points will only be granted by Brainly if the answer is applicable.)

Answer 1

Answer:

equation is

x2(a2 + b2)+ 2x (ac + bd) +(c2 + d2)= 0 .

Let D be the discriminant of this equation.

Therefore,D = 4(ac + bd)2 - 4(a2 + b2)(c2 + d2)

⇒ D = 4 [(ac+ bd)2 - (a2 + b2) (c2 + d2)]

⇒ D = 4 [a2c2 + b2d2 + 2ac.bd - a2c2 - a2d2 - b2c2 - b2d2]

⇒D = 4 [2ac.bd - a2d2 - b2c2] = - 4[a2d2 + b2c2 - 2ad.bc] = -4(ad - bc)2

It is given that ad ≠ bc.

⇒ ad - bc ≠ 0

⇒ (ad - bc)2 > 0

⇒ - 4 (ad - bc)2 < 0

⇒ D < 0.

Therefore, given equation has no real root.

Answer 2

Define vectors p and q in cartesian coordinates:

p⃗ =(a,b,0)

q⃗ =(c,d,0)

and let θ be the angle between p⃗ and q⃗ .

Then

1=sin2θ+ cos2θ

p2q2= p2q2 sin2θ+ p2q2 cos2θ

|p⃗ |2|q⃗ |2=( pq sinθ)2+( pq cosθ)2

|p⃗ |2|q⃗ |2=(p⃗ ⋅q⃗ )2+|p⃗ ×q⃗ |2

(a2+b2)(c2+d2)=(ac+bd )2+(ad−bc )2