Question 17683
If {{{p*(q-r)*x^2+q*(r-p)*x*y+r*(p-q)*y^2}}} is a perfect square, then prove p,q,r are in harmonic progression.

I know that i should prove somehow that {{{q=(2*p*r)/(p+r)}}} i think i can do this using {{{b^2=4*a*c}}} but i thought that it only applied to quadratic equations of the form {{{a*x^2+b*x+c}}} Please help. Thanks
YOU ARE PERFECTLY CORRECT...PROCEED AND YOU WOULD HAVE GOT THE ANSWER BY YOUR SELF.DONT WORRY ABOUT PRESENCE OF CONSTANTS LIKE A,B,C OR VARIABLES LIKE Y,Y^2 ETC.IN THEIR PLACE..IT IS ALL SAME AS LONG AS YOU HAVE A QUADRATIC IN ONE VARIABLE.IT COULD BE IN X OR Y.
{{{b^2=4*a*c}}} =0 for the given equation to be a perfect square..we have 
a=p(q-r)..b=q(r-p)y...c=r*(p-q)*y^2..so,we have
b^2=4ac ...hence...
q^2*(r-p)^2*y^2=4*p*(q-r)*r*(p-q)*y^2
q^2*(r-p)^2=4*p*(q-r)*r*(p-q)
..NOW HERE LET US USE A SMALL TRICK TO SIMPLIFY THINGS OR DO IT ORALLY....
LET t=p(q-r)...u=q(r-p)....v=r(p-q).so that.t+u+v=0.Or.-u=(t+v)..or.u^2=(t+v)^2
but u^2=4tv from above q^2*(r-p)^2=4*p*(q-r)*r*(p-q)
hence (t+v)^2=4tv..or (t-v)^2=0 ..or t=v..or p(q-r)=r(p-q).
NOW YOU HAVE YOUR DESIRED EQUATION 
{{{q=(2*p*r)/(p+r)}}} and hence p,q,r are in H.P.