If the sum of the roots of the equation ax2 + bx + c=0 is equal to the sum of the squares of their reciprocals

Question

then a/c, b/a, c/b are in

Answer 1

We assume a, b, c are not 0, else it makes no sense to talk about a/c, b/a and c/b.

Suppose the roots are r and s. Then (x-r)(x-s) = 0, so
x^2 - (r+s) x + (rs) = 0
and hence r+s = -b/a and rs = c/a.

We are saying that
r + s = (1/r^2 + 1/s^2)
<=> (rs)^2 (r+s) = (rs)^2 (1/r^2 + 1/s^2) = s^2 + r^2
<=> (rs)^2 (r+s) = (r+s)^2 - 2rs
<=> (c/a)^2 (-b/a) = (-b/a)^2 - 2(c/a)
Multiplying through by a^3 we get
-bc^2 = ab^2 - 2a^2c
Divide through by abc to get
-c/a = b/c - 2a/b
<=> a/b - c/a = b/c - a/b
which implies c/a, a/b and b/c are in arithmetic progression.

Hence their reciprocals, a/c, b/a and c/b are in harmonic progression.