Question

prove that the equation x2 ( a2 + b2 ) + 2x ( ac + bd ) + ( c2 +d2 ) =0 has no real roots if ad=bc(ad not equal to bc).

Answer 1

Compare the given quadratic equation with Ax2 + Bx + C = 0

Here A = a2 + b2 , B = 2(ac + bd) and C = c2 + d2

Consider, B2 – 4AC = [2(ac + bd)]2 – 4 ´ (a2 + b2 ) x (c2 +d2)

= 4[a2c2 + 2abcd + b2d2] – 4[a2c2 + a2d2 + b2c2 + b2d2]

= 4a2c2 + 8abcd + 4b2d2 – 4a2c2 – 4a2d2 – 4b2c2 – 4b2d2

= 8abcd– 4a2d2 – 4b2c2

= –4[4a2d2 + 4b2c2 – 2abcd]

= –4[ad – bc]2

Hence the given equation has no real roots unless ad ¹ bc

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let D be the discriminant of the equation  (a² + b²)x² + 2x(ac + bd) + (c² + d²) = 0

Then,

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

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

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

⇒ D = 4[2ac × bd - a²d² - b²c²]

⇒ D = - 4[a²d² + b²c² - 2ad × bc]

⇒ D = - 4(ad - bc)²

It is given that ad ≠ bc

⇒ ad - bc ≠ 0

⇒ (ad - bc)² > 0

⇒ - 4(ad - bc)² < 0

⇒ D < 0

Hence, the given equation has no real roots.