SOLUTION: one of the root of the equation x^2-px+q=0 is the square of the other.proof that pq-q(3p+1)-q^2=0

I'm sorry but what you are asked to prove is
simply not true.  Here is why:

Take the equation

x² - 6x + 8 = 0

Solving it by factoring the left side:

(x - 2)(x - 4) = 0

x - 2 = 0;  x - 4 = 0

    x = 2;      x = 4

So one root is 2 and the other root is 4

Since 4 is the square of 2, the equation

x² - 6x + 8 = 0 fulfills the requirements that

one of the roots is the square of the other.  

So if we compare it to:

x² - px + q = 0

We see that p=6 and q=8.

So if what you are asked to prove were true, then
this 

  pq-q(3p+1)-q² = 0 

would have to be true when we substitute p=6 and q=8

Substituting:

6·8-8(3·6+1)-8² = 0
  48-8(18+1)-64 = 0
    48-8(19)-64 = 0
      48-152-64 = 0
           -168 = 0

But as you see, this is not true at all.  So the 
problem is botched.  If you can correct it in
the thank-you note, we will help you.

Edwin