if $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has repeated roots prove... if the equation $$a(b-c)x^2+b(c-a)x+c(a-b)=0$$ has repeated roots prove that $${1\over a}, {1\over b},{1\over c} $$  are in Arithmetic Progression
Any idea about how to go about solving this ? Thanks is advance!

With $a,b,c\ne 0$, observe that $1/a,1/b,1/c$ is an Arithmetic Progression iff $$K=0\text {, where}$$ $$K=a b+b c-2a c.$$ The condition that the quadratic has a repeated root is that the discriminant is zero. We have $$0=b^2(c-a)^2-4a c(a-b)(b-c)\iff$$ $$0=b^2c^2+b^2a^2-2b^2a c-4a c(-b^2+b c+b a-ac)\iff$$  $$0=b^2c^2+b^2a^2+2a c b^2-4a c(b c+b a-a c)\iff$$ $$0= b^2c^2+b^2a^2+2a c b^2-4 a c(b c+b a-2a c)-4a^2c^2\iff$$ $$0=b^2c^2+b^2a^2+2a c b^2-4a c K-4a^2c^2\iff$$ $$0=(b c+b a)^2-4a c K-4a^2c^2\iff$$  $$0=(b c+b a)^2 -(2a c)^2-4a c K\iff$$ $$0=(b c+b a-2a c)(b c+b a+2a c)-4a c K$$ $$0=K(b c+b a+2a c)-4a c K\iff$$ $$0=K(b c+b a-2a c)\iff$$  $$0=K^2.$$
