In this approach we don't actually find the roots as the other
tutor did, although that method is quite correct.

x² + px + q = 0

Let the roots be r1 and r2.  
Then the above factors as

(x - r1)(x - r2) = 0

or, upon "FOIL"ing that out:

x² - r2x - r1x + r1r2 = 0

x² - (r2+r1)x + r1r2 = 0

x² - (r1+r2)x + r1r2 = 0

comparing this to

x² + px + q = 0

p = -(r1+r2)

q = r1r2

Now we use the fact that the sum of its roots 
are equal to thrice their difference, which
says:

r1+r2 = 3(r1-r2)  or r1+r2 = 3(r2-r1)

Both equations are technically necessary because we are not told 
which order we are to subtract the roots to find the difference.

r1+r2 = 3r1-3r2  or r1+r2 = 3r2-3r1
  4r2 = 2r1      or   4r1 = 2r2
  2r2 = r1       or   2r1 = r2

But since they are the same except for which root we call
r1 and which we call r2, we can just take the first case.

Now we have these 3 equations:

p = -(r1+r2)
q = r1r2
2r2 = r1

Using the third equation, we substitute 2r2 for r1 in the
first two equations:

p = -(2r2+r2)
q = 2r2r2

Simplifying,

p = -3r2
q = 2r2²

Solve the first one for r2:

r2 = 
r2 = 

Substitute  for r2 in q = 2r2²:

q = 2
q = 2
q = 2

Multiply both sides by 9

9q = 2p² 

That's the same as

2p² = 9q

Edwin