Question

a=a(b-c),b=b(c-a),c=c(a-b) then solve b^2-4ac=0.​

Answer 1

. let r & s be the roots, then: r + s = -(b - c)/(a - b) but r = s: 2r = -(b - c)/(a - b) r = -(b - c )/[2(a - b)] also: r * s = r^2 = (c - a)/(a - b)