Question

Plz answer fast => If the roots of the equation (b-c)x²+(c-a)x+(a-b) = 0

then prove that 2b = a+b

Answer 1

 For the quadratic equation ax² + bx + c = 0, 
the discriminant Δ = b² - 4ac = 0 for two equal roots 

So for the quadratic equation (b-c)x^2 +( c-a)x + (a-b) = 0 to have two equal roots:  (c-a)² - 4(b-c)(a-b)= 0 

The given equation is (b-c)x²+(c-a)x+(a-b) = 0

⇒ c² - 2ac + a² - 4ab + 4b² + 4ac - 4bc = 0 


⇒ c² + 2ac + a² + 4b² - 4ab - 4bc = 0 


⇒ (c + a)² + 4b² - 4b(a + c) = 0 


⇒ (c + a)² - 4b(a + c) + (2b)² = 0 


⇒ [(a + c) - (2b)]² = 0 


 a + c = 2b QED