Question

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

Answer 1

(b-c)x² + x(c-a) + (a-b) =0
To have real and equal roots, Discriminant, D =0
(c-a)² - 4(a-b)(b-c) =0
⇒c² +a² - 2ac - 4(ab -ac -b²+ bc) =0
⇒c² + a² -2ac - 4ab +4ac +4b² -4bc =0
⇒a² +4b² +c² - 4ab - 4bc +2ac =0
⇒(a -2b +c)² =0
∴(a -2b +c) =0
⇒(a+c) =2b   (proved )
a,b and c are in A.P , where b is A.M

Answer 2

Some intelligency is required to solve this question smoothly. So please have a look on the above solution and practice it simultaneously. You will not have any fear in these types of questions anymore on doing so.