Question

If the roots of the equation a (b –

c.x2 + b (c –

a.x + c (a –

b.= 0 are equal, then show that 1/a , 1/b , 1/c are in

a.p.

Answer 1

plz mark as brainlest answer

sol:

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

Here a = b - c , b = b (c – a) , c = c (a – b)

If roots are real and equal then b2 - 4ac = 0

⇒ [b (c – a)]2 - 4(b - c)c (a – b) = 0

⇒ b2[ c2 + a2 - 2ac] - 4[ ab - b2 - ac + bc] c = 0

⇒ [b2c2 + a2b2 - 2ab2c] - 4[ abc - b2c - ac2 + bc2] = 0

⇒ b2c2 + a2b2 - 2ab2c - 4abc + 4b2c + 4ac2 - 4bc2 = 0

simplify we get 1/a , 1/b , 1 / c are in AP.