Question

If the roots of the equation (b − c) x² + (c − a) x + (a − b) = 0 are equal, then prove that 2b = a + c.

Answer 1

The given quadric equation is (b − c) x² + (c − a) x + (a − b) = 0, and roots are real

Then prove that 2b = a + c

Here,

a = (b − c), b = (c − a) and c = (a − b)

As we know that D = b² − 4ac

Putting the value of a = (b − c), b = (c − a) and c = (a − b)

D = b² − 4ac

= (c − a)² − 4 × (b − c) × (a − b)

= c² − 2ca + a² − 4 (ab − b² − ca + bc)

= c² − 2ca + a² − 4ab + 4b² + 4ca − 4bc

= c² + 2ca + a² − 4ab + 4b² − 4bc

= a² + 4b² + c² + 2ca − 4ab − 4bc

As we know that (a² + 4b² + c² + 2ca − 4ab − 4bc) = (a + c − 2b)²

D = (a + c − 2b)²

The given equation will have real roots, if D = 0

(a + c − 2b)² = 0

Square root both side we get

`sqrt((a + c - 2b)^2)=0`

a + c − 2b = 0

a + c = 2b

Hence 2b = a + c.