Question

If the root's of the equation (b-c)x^2+(c-a)x+(a+b)=0are equal then prove that 2b=a+c

Answer 1

if the roots are equal
...simply D=0...therefore b^2-4ac=0. simply replace a b and c from the given quadratic and you will get it

Answer 2

Answer:

we know that if the roots of the Quadratic equation are equal then the ∆ = b² - 4ac = 0

(c-a)² - 4 (b-c) (a-b) = 0

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

c² + a² + 2ac - 4ab + 4b² - 4bc = 0

c² + (-2b)² + a² + 2(a)(-2b) + 2(-2b)(c) + 2(a)(c) = 0

• [ x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)² ]

(c + (-2b) + a)² = 0

c - 2b + a = 0

2b = a + c

Hope it's helpful to you and all