Question

If the roots of tge equation (a^2+b^2)x^2+2 (bc-ad)x+c^2+d^2=0 are real and equal show that ac+bd =0


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Answer 1

The roots are real and equql.

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Answer 2

Answer:

ac + bd = 0

Step-by-step explanation:

Given Equation is (a² + b²)x² + 2(bc - ad)x + c² + d² = 0.

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

a = (a² + b²), b = 2(bc - ad), c = c² + d².

Now,

Given that the equation has real and equal roots.

⇒ b² - 4ac = 0

⇒ [2(bc - ad)]² - 4(a² + b²)(c² + d²) = 0

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

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

⇒ -4b²c² - 4a²c² - 8bcad = 0

⇒ -4(a²c² + b²c² + 2abcd) = 0

⇒ (ac + bd)² = 0

⇒ ac + bd = 0.

Hope it helps!