If roots of quadratic equation (b – c)x2 + (c – a)x + (a – b)

1.	If roots of quadratic equation (b – c)x2 + (c – a)x + (a – b) = 0 are real and equal, then prove that 2b = a + c
Answer» Given equation is (b – c)x² + (c – a)x + (a – b) = 0 Comparing it by general quadratic equation, a = (b – c), b = (c – a) and c = (a – b) Discriminant (D) = b² – 4ac = (c – a)² – 4(b – c)(a – b) = (c² + a² – 2ac) – 4(ab – ac – b² + bc) = c² + a² – 2ac – 4ab + 4ac + 4b² – 4bc = c² + a² + 2ac – 4ab – 4bc + 4b² = (a + c)² – 4 b(a + c) + 4b² = (a + c)² – 2 × (a + c) × 2b + (2b)² = [(a + c) – 2b]² Roots of equation are real and equal. D = 0 ⇒ [(a + c) – 2b]² = 0 ⇒ (a + c) – 2b = 0 ⇒ a + c = 2b

If roots of quadratic equation (b – c)x2 + (c – a)x + (a – b) = 0 are real and equal, then prove that 2b = a + c

Answer»

Given equation is (b – c)x² + (c – a)x + (a – b) = 0

Comparing it by general quadratic equation, a = (b – c), b = (c – a) and c = (a – b)

Discriminant (D) = b² – 4ac

= (c – a)² – 4(b – c)(a – b)

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

= c² + a² – 2ac – 4ab + 4ac + 4b² – 4bc

= c² + a² + 2ac – 4ab – 4bc + 4b²

= (a + c)² – 4 b(a + c) + 4b²

= (a + c)² – 2 × (a + c) × 2b + (2b)²

= [(a + c) – 2b]²

Roots of equation are real and equal.

D = 0

⇒ [(a + c) – 2b]² = 0

⇒ (a + c) – 2b = 0

⇒ a + c = 2b