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 x (b - c) x (c - a)

= 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.