Understanding 2012 AMC 12B #23 
Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that
  $P(Q(x))$ has zeros at $x=-23$, $-21$, $-17$, and $-15$, and $Q(P(x))$ has zeros
  at $x=-59$,$-57$,$-51$ and $-49$. What is the sum of the minimum values of
  $P(x)$ and $Q(x)$?

I found a solution here, and I was able to follow it up until this sentence:

Since Q is monic, the roots of $Q(x) = a + \sqrt {b}$ are "farther"
  from the axis of symmetry than the roots of $Q(x) = a - \sqrt {b}$. 
Thus, we have
$Q( - 23) = - 54 + \sqrt {b}$
$Q( -21) =- 54 - \sqrt {b},$
or substituting
$16 - d = - 54 + \sqrt {b}$
$ 4 - d = - 54 - \sqrt {b}$

What does it mean by the roots are farther from the axis of symmetry, and how did they arrive at that? And from that, how does one get the mentioned equations?

Claim: If $ \alpha, \beta$ are the roots of $ Q(x) = a + \sqrt{b}$ and $\gamma , \delta$ are the roots of $Q(x) = a - \sqrt{b}$, then
$$|\alpha - \beta | > | \gamma - \delta|.$$

 Proof: This follows by considering the graph of the monic quadratic $y=Q(x)$ and the lines $ y = a + \sqrt{b}$ and $ y = a - \sqrt{b}$. Use the fact that $ a + \sqrt{b} > a - \sqrt{b}$. $_\square$

The axis of symmetry is the line $ x = \frac{\alpha + \beta} { 2} = \frac{ \gamma + \delta} {2}$. Hence, from the above inequality $\alpha, \beta$ are further away from this axis, than $\gamma, \delta$.
