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

Algebra ->  Polynomials-and-rational-expressions -> 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      Log On


   

Question 17683: Hi i would really appreciate it if someone would help me with this problem.
If p%2A%28q-r%29%2Ax%5E2%2Bq%2A%28r-p%29%2Ax%2Ay%2Br%2A%28p-q%29%2Ay%5E2 is a perfect square, then prove p,q,r are in harmonic progression.
I know that i should prove somehow that q=%282%2Ap%2Ar%29%2F%28p%2Br%29 i think i can do this using b%5E2=4%2Aa%2Ac but i thought that it only applied to quadratic equations of the form a%2Ax%5E2%2Bb%2Ax%2Bc Please help. Thanks
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
If p%2A%28q-r%29%2Ax%5E2%2Bq%2A%28r-p%29%2Ax%2Ay%2Br%2A%28p-q%29%2Ay%5E2 is a perfect square, then prove p,q,r are in harmonic progression.
I know that i should prove somehow that q=%282%2Ap%2Ar%29%2F%28p%2Br%29 i think i can do this using b%5E2=4%2Aa%2Ac but i thought that it only applied to quadratic equations of the form a%2Ax%5E2%2Bb%2Ax%2Bc 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%5E2=4%2Aa%2Ac =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=%282%2Ap%2Ar%29%2F%28p%2Br%29 and hence p,q,r are in H.P.