Question

If the roots of the quadratic equation ax^2+bx+c=0 are in the ratio r:1 prove that
acr^2+(2ac_b^2)r+ac=0

Answer 1

If the roots are p and rp, then you have

b/a = -(p+rp)
c/a = p*rp

see what you can do with that

Answer 2

let the roots of ax^2+bx+c=0 be m and n
then
m:n = r:1
m/n =r/1
m = rn

sum of roots = m+n = rn + n = n(r+1) = -b/a **
product of roots = mn = n^2 r = c/a ***

divide ** by ***
(r+1)/nr = (-b/a) / (c/a) = -b/c
rc + c = -bnr
rc + bnr = -c
r(c + bn) = -c
r = -c/(c+bn)

For acr^2+(2ac - b^2)r+ac=0
LS = ac(-c/(c+bn) )^2 + (2ac - b^2)(-c/(c+bn)) + ac
= -ac^3/(c + bn)^2 -c(2ac - b^2)/(c+bn) + ac

I can't get this to be zero, as it should
and I can't find my error.
perhaps you can, I am pretty sure of my method.

Answer 3

Take a look here, and all will become clear.

https://in.answers.yahoo.com/question/index?qid=20130823004513AAdmLmh