Question

If the roots the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that b + c = 2a.

Answer 1

Given equation is (a - b)x² + (b - c)x + (c - a) = 0
It is of the form Ax² + Bx + C = 0
We know that D = B² - 4AC
∴ D = (b - c)² - 4(a - b)(c - a) = 0
For real and equal roots, D = 0
Now, D = 0
⇒ (b - c)² - 4(a - b)(c - a) = 0
⇒b² - 2bc + c² - 4(ac - a² - bc + ab ) = 0
⇒b² - 2bc + c² - 4ac + 4a² + 4bc - 4ab = 0
⇒ 4a² + b² + c² - 4ab + 2bc - 4ac = 0
⇒ (-2a)² + b² + c² + 2(-2a)b + 2bc + 2c(- 2a) = 0
⇒ [ (- 2a ) + b + c]² = 0
⇒ (- 2a + b + c)² = 0
⇒ - 2a + b + c = 0
⇒ b + c = 2a