Let a and b be the roots of the equation. Hence,

`a^2 + ap + q = 0` and

`b^2 +pb + q = 0`

We know that `a + b = a^2 + b^2` . Adding the two equations yield:``

`a^2 + ap + q + b^2 + pb +...

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Let a and b be the roots of the equation. Hence,

`a^2 + ap + q = 0` and

`b^2 +pb + q = 0`

We know that `a + b = a^2 + b^2` . Adding the two equations yield:``

`a^2 + ap + q + b^2 + pb + q = 0`

Hence,

`a^2 + b^2 + (a + b)p + 2q = 0`

But `a + b = a^2 + b^2` and we simply let this sum be represented by `S` such that:

`S + Sp + 2q = 0`

Manipulating the equation (to make it look like what we want to prove):

`-S-Sp = 2q`

`-S ( p + 1) = 2q`

Using Viete's formula, we know that the sum of the roots of an equation is just the negative of the coefficient of `x` , in this case:

`a + b = -p`

We know that `S = a + b = a^2 + b^2` . Hence:

`-(-p)(p+1) = 2q`

`p (p+1) = 2q` QED

(Note that I only used S for compactness of expression, and since `a+b = a^2 + b^2` I just used it to represent both.)