Question 696524
<pre>
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 r<sub>1</sub> and r<sub>2</sub>.  
Then the above factors as

(x - r<sub>1</sub>)(x - r<sub>2</sub>) = 0

or, upon "FOIL"ing that out:

x² - r<sub>2</sub>x - r<sub>1</sub>x + r<sub>1</sub>r<sub>2</sub> = 0

x² - (r<sub>2</sub>+r<sub>1</sub>)x + r<sub>1</sub>r<sub>2</sub> = 0

x² - (r<sub>1</sub>+r<sub>2</sub>)x + r<sub>1</sub>r<sub>2</sub> = 0

comparing this to

x² + px + q = 0

p = -(r<sub>1</sub>+r<sub>2</sub>)

q = r<sub>1</sub>r<sub>2</sub>

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

r<sub>1</sub>+r<sub>2</sub> = 3(r<sub>1</sub>-r<sub>2</sub>)  or r<sub>1</sub>+r<sub>2</sub> = 3(r<sub>2</sub>-r<sub>1</sub>)

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

r<sub>1</sub>+r<sub>2</sub> = 3r<sub>1</sub>-3r<sub>2</sub>  or r<sub>1</sub>+r<sub>2</sub> = 3r<sub>2</sub>-3r<sub>1</sub>
  4r<sub>2</sub> = 2r<sub>1</sub>      or   4r<sub>1</sub> = 2r<sub>2</sub>
  2r<sub>2</sub> = r<sub>1</sub>       or   2r<sub>1</sub> = r<sub>2</sub>

But since they are the same except for which root we call
r<sub>1</sub> and which we call r<sub>2</sub>, we can just take the first case.

Now we have these 3 equations:

p = -(r<sub>1</sub>+r<sub>2</sub>)
q = r<sub>1</sub>r<sub>2</sub>
2r<sub>2</sub> = r<sub>1</sub>

Using the third equation, we substitute 2r<sub>2</sub> for r<sub>1</sub> in the
first two equations:

p = -(2r<sub>2</sub>+r<sub>2</sub>)
q = 2r<sub>2</sub>r<sub>2</sub>

Simplifying,

p = -3r<sub>2</sub>
q = 2r<sub>2</sub>²

Solve the first one for r<sub>2</sub>:

r<sub>2</sub> = {{{p/(-3)}}}
r<sub>2</sub> = {{{-p/3}}}

Substitute {{{-p/3}}} for r<sub>2</sub> in q = 2r<sub>2</sub>²:

q = 2{{{(-p/3)^2}}}
q = 2{{{(p^2/3^2)}}}
q = 2{{{(p^2/9)}}}

Multiply both sides by 9

9q = 2p² 

That's the same as

2p² = 9q

Edwin</pre>