Question

if the roots of quadratic equation (a-b)x²+(b-c)x+(c-a)=0 are equal, then prove that 2a=b+c​

Answer 1

Please see the attachment

Answer 2

(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