Question

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

Answer 1

If roots of a quadratic equation are equal, then discriminant of the quadratic equation is 0

D = b² − 4ac = 0

(b−c)x² + (c−a)x + (a−b) = 0

Comparing with 

ax² + bx + c = 0

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

So,

Answer 2

If roots of a quadratic equation are equal,

then discriminant of the quadratic equation is 0

D=b²−4ac=0

(b−c)x²+(c−a)x+(a−b)=0

Comparing with

ax²+bx+c=0

Here, a=(b−c), b=(c−a) and c=(a−b)

So,

(c−a)²−4(b−c)(a−b)=0

c²+a²−2ac−4(ab−b²−ac+bc)=0

c²+a²−2ac−4ab+4b²+4ac−4bc=0

c²+a²+2ac+4b²−4ab−4bc=0

(c+a)²+4b²−4b(a+c)=0

(c+a)²+(2b)²−2(c+a)(2b)=0

[(c+a)−(2b)]²=0

c+a−2b=0

2b=c+a

Hence Proved