Question

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

Answer 1

(b-c)x²+(c-a)x+(a-b)=0

Comparing with quadratic equation

Ax²+Bx+C=0

A=(b-c),B=c-a,C=a-b

When roots are equal ,its discriminant will be zero,means√( b^2-4ac)=0 i.e

b^2-4ac=0

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b

LIKE IF SATISFIED