Question

If the roots of the equation (b-c) x + (c-a) x + (a-b) =0 are equal, then prove that 2b = a+c
1​

Answer 1

Step-by-step explanation:

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

D=b^2−4ac=0

(b−c)x^2+(c−a)x+(a−b)=0

Comparing with  ax ^2+bx+c=0

Here, a=(b−c), b=(c−a) and c=(a−b)

So,

⇒(c−a) ^2 −4(b−c)(a−b)=0

⇒c^2+a^2−2ac−4(ab−b^2−ac+bc)=0

⇒c^2+a ^2−2ac−4ab+4b ^2+4ac−4bc=0

⇒c^2 +a^2 +2ac+4b ^2−4ab−4bc=0

⇒(c+a) ^2+4b ^2 −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