Question

Please answer q14
Pleae tell how to proceed in the question

Answer 1

(b-c)x^2 +(c-a)x+(a-b)=0
$d = b {}^{2} - 4ac$
=(c-a)^2-4((b-c)(a-b))
=c^2-2ac+a^2-4(ab-ac-b^2+bc)
=c^2-2ac+a^2-4ab+4ac+4b^2-4bc
=a^2+4b^2+c^2+2ac-4ab-4bc
=(a-2b+c)^2=0
=(a-2b+c)=0
=a+c=2b
hence proved