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

1.	If the roots of the equation a(b – c) x2 + b(c – a)x + c(a – b) = 0 are equal, then a, b, c are in :(a) AP (b) GP (c) HP (d) None of these
Answer» (c) HP If the roots of the equation a(b – c)x² + b(c – a)x + c(a – b) = 0 are equal, then Discriminant (D) = 0, i.e., ⇒ b² (c – a)² – 4a(b – c) c(a – b) = 0. ⇒ b² (c² + a² – 2ac) – 4ac (ab – ca – b² + bc) = 0 ⇒ b²c² + b² a² – 2ab²c – 4a²bc + 4a²c² + 4ab²c – 4abc² = 0 ⇒ a²b² + b²c² + 4a²c² + 2ab²c – 4a²bc – 4abc² = 0 ⇒ (ab + bc – 2ac)² = 0 ⇒ ab + bc – 2ac = 0 ⇒ ab + bc = 2ac ⇒ \(\frac{1}{c}+\frac{1}{a}=\frac{2}{b}\) ⇒ a, b, c are in H.P.

If the roots of the equation a(b – c) x2 + b(c – a)x + c(a – b) = 0 are equal, then a, b, c are in :(a) AP (b) GP (c) HP (d) None of these

Answer»

(c) HP

If the roots of the equation a(b – c)x² + b(c – a)x + c(a – b) = 0 are equal, then Discriminant (D) = 0, i.e.,

⇒ b² (c – a)² – 4a(b – c) c(a – b) = 0.

⇒ b² (c² + a² – 2ac) – 4ac (ab – ca – b² + bc) = 0

⇒ b²c² + b² a² – 2ab²c – 4a²bc + 4a²c² + 4ab²c – 4abc² = 0

⇒ a²b² + b²c² + 4a²c² + 2ab²c – 4a²bc – 4abc² = 0

⇒ (ab + bc – 2ac)² = 0 ⇒ ab + bc – 2ac = 0

⇒ ab + bc = 2ac ⇒ \(\frac{1}{c}+\frac{1}{a}=\frac{2}{b}\) ⇒ a, b, c are in H.P.