SOLUTION: Hi i would really appreciate it if someone would help me with this problem. 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 pr

Click here to see ALL problems on Polynomials-and-rational-expressions

Question 17683: Hi i would really appreciate it if someone would help me with this problem.
If is a perfect square, then prove p,q,r are in harmonic progression.
I know that i should prove somehow that i think i can do this using but i thought that it only applied to quadratic equations of the form Please help. Thanks
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
If is a perfect square, then prove p,q,r are in harmonic progression.
I know that i should prove somehow that i think i can do this using but i thought that it only applied to quadratic equations of the form 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.
=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
and hence p,q,r are in H.P.