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Answer 1

Answer:

Option - (D)

Step-by-step explanation:

Given Equation is (a² + b²)t² - 2(ac + bd)t + (c² + d²) = 0.

On comparing with ax² + bx + c = 0, we get

a = (a² + b²), b = -2(ac + bd), c = (c² + d²)

Given that the equation has real roots.

∴ Δ = 0

⇒ b² - 4ac = 0

⇒ (-2(ac + bd))² - 4(a² + b²)(c² + d²) = 0

⇒ 4(ac + bd)² - 4(a²c² + a²d² + b²c² + b²d²) = 0

⇒ 4(a²c² + b²d² + 2acbd) - 4a²c² - 4a²d² - 4b²c² - 4b²d² = 0

⇒ 4a²c² + 4b²d² + 8abcd - 4a²c² - 4a²d² - 4b²c² - 4b²d² = 0

⇒ 8abcd - 4a²d² - 4b²c² = 0

⇒ 8abcd = 4a²d² + 4b²c²

⇒ 8abcd = 4(a²d² + b²c²)

⇒ 2abcd = a²d² + b²c²

⇒ 2 = (a²d²/abcd) + (b²c²/abcd)

⇒ 2 = ad/bc + bc/ad

⇒ 2bcad = a²d² + b²c²

⇒ a²d² + b²c² - 2adbc = 0

⇒ (ad - bc) = 0

⇒ ad = bc

⇒ (a/b) = (c/d)

Therefore, the answer is Option(D).

Hope this helps!

Answer 2

Answer:

Option - D

Step-by-step explanation:

Given Equation is (a² + b²)t² - 2(ac + bd)t + (c² + d²) = 0.

On comparing with ax² + bx + c = 0, we get

a = (a² + b²), b = -2(ac + bd), c = (c² + d²)

Given that the equation has real roots.

Δ = 0

b² - 4ac = 0

(-2(ac + bd))² - 4(a² + b²)(c² + d²) = 0

4(ac + bd)² - 4(a²c² + a²d² + b²c² + b²d²) = 0

4(a²c² + b²d² + 2acbd) - 4a²c² - 4a²d² - 4b²c² - 4b²d² = 0

4a²c² + 4b²d² + 8abcd - 4a²c² - 4a²d² - 4b²c² - 4b²d² = 0

8abcd - 4a²d² - 4b²c² = 0

8abcd = 4a²d² + 4b²c²

8abcd = 4(a²d² + b²c²)

2abcd = a²d² + b²c²

2 = (a²d²/abcd) + (b²c²/abcd)

2 = ad/bc + bc/ad

2bcad = a²d² + b²c²

a²d² + b²c² - 2adbc = 0

(ad - bc) = 0

ad = bc

(a/b) = (c/d)