If the roots of the equation (b - c)x^{2} + (c - a)x + (a - b) = 0 are equal, then prove that 2b = a + c.

#### Solution

The given quadric equation is(b - c)x^{2} + (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 = b^{2} - 4ac

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

D = b^{2} - 4ac

= (c - a)^{2} - 4 x (b - c) x (c - a)

= c^{2} - 2ca + a^{2} - 4 (ab - b^{2} - ca + bc)

= c^{2} - 2ca + a^{2} - 4ab + 4b^{2} + 4ca - 4bc

= c^{2} + 2ca + a^{2} - 4ab + 4b^{2} - 4bc

= a^{2} + 4b^{2} + c^{2} + 2ca - 4ab - 4bc

As we know that (a^{2} + 4b^{2} + c^{2} + 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.