If a,b,c are in Geometric Progression, then the equations $ax^2+2bx+c=0$ and $dx2+2ex+f=0$ have a common root if $\frac da, \frac eb, \frac fc$ are in:
- Arithmetic Progression
- Geometric Progression
- Harmonic Progression
Since this question was asked already and many simple approach has been mentioned in the answers but I wanted to solve this question through common root condition only. So I tried the condition for one common root between two quadratic equation
$(a_1b_2−b_1a_2)(b_1c_2−c_1b_2)=(c_1a_2−a_1c_2)^2$
substituting the coefficient:
$(2ae−2bd)(2bf-2ce) = (cd−af)^2$ ........(I)
and also also substituted the value of b and c with $ar$ and $ar^2$ respectively as mentioned in the answers but I couldn't simplify eqn (I) to
$\frac da + \frac fc = \frac {2e}{b}$. which would help in proving that they are in A.P.
Can anybody help me with the simplification of the equation?
Edit1:
Solution after suggestion from Blue:
$(2ae−2bd)(2bf-2ce) = (cd−af)^2$
$u \to d/a $
$v \to e/b $
$w \to c/f $
$(2ae−2bd)(2bf-2ce) = (cd−af)^2$
$(2abv−2bau)(2bcw-2bcv) = (acu−acw)^2$
$4ab.bc.(v−u)(w-v) = (ac)^2(u−w)^2$
$4ab^2c(v−u)(w-v) = (ac)^2(u−w)^2$
$4b^2(v−u)(w-v) = ac(u−w)^2$
as $abc$ are in G.P
$b^2 \to ac $
$4b^2(v−u)(w-v) = b^2(u−w)^2$
$4(v−u)(w-v) = (u−w)^2$
$4vw−4v^2-4uw+4uv = u^2+w^2-2uw$
$ u^2+w^2+(2v)^2+2uw-4vw-4uv=0$
$ (u+w-2v)^2=0$
$ u+w-2v=0$
$ u+w=2v$