If the roots of the equation a(b – c)x2 + b(c – a)x + c(a – b) = 0 are equal, show that 2/b=1/a+1/c.
(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.
Since the roots of the given equations are equal, so discriminant will be equal to zero.
=> b2(c – a)2 – 4a(b – c) . c(a – b) = 0
=> b2(c2 + a2 – 2ac) – 4ac(ba – ca – b2 + bc) = 0,
=> a2b2 + b2c2 + 4a2c2 + 2b2ac – 4ac2bc – 4abc2 = 0
=> (ab + bc – 2ac)2 = 0
=> ab + bc – 2ac = 0
=> ab + bc = 2ac
=> 1/c+1/a=2/b
=> 2/b=1/a+1/c.
Hence Proved.