<s>[INST] Prove that if $k^2+pk+q=0$ and $l^2-pl-q=0$ then between $k$ and $l$ there is a solution to $x^2-2px-2q=0$. Real number $k$ satisfies $x^2+px+q=0$ and $l$ satisfies $x^2-px-q=0$. Prove that between $k$ and $l$ there is a solution to the equation $x^2-2px-2q=0$.
I started the solution this way:
$$ k = \frac{-p\pm\sqrt{p^2-4q}}{2} $$
$$ l = \frac{p\pm\sqrt{p^2+4q}}{2} $$
$$ x = p\pm\sqrt{p^2+2q} $$
But I failed to compare these solutions. What should be done next?
 [/INST] The comparison argument you mentioned might run something like this.  As Dhanvi Sreenivasan notes, all three parabolas for the quadratic equations have a mutual intersection at $ \ x = -\frac{q}{p} \ \ . $  If we inspect a graph, the zeroes
$$ k  \ =  \ \frac{-p}{2} \ + \ \frac{\sqrt{p^2 \ - \ 4q}}{2} \ \ ,  \ \  l \ =  \ \frac{p}{2} \ - \ \frac{\sqrt{p^2 \ + \ 4q}}{2} \ \ ,  \ \ \text{and} \ \ r \ = \ p \ - \ \sqrt{p^2 \ + \ 2q} $$
appear to lie in the order $ \ k < r < l \ $ , so we will attempt to prove that $ \ r \ $ always lies in the interval $ \ [k \ , \ l] \ \ . $    The condition for the proposition to be meaningful is that for $ \ q > 0 \ \ , $ we must have $ \ p^2 > 4q \ \ , $ and for $ \ q < 0 \ \ , \ \ p^2 > -4q \ \ . $  This can be summarized as $ \ p^2 > 4·|q| \ \ ; $ however, since the situation of the parabolas is symmetrical about the $ \ y-$axis for these two cases, we only need to consider $ \ q > 0 \ \ . $  It is also clear that it will suffice to investigate $ \ p > 0 \ \ , $ as taking $ \ p < 0 \ $ produces a mirror-image arrangement of the geometrical situation.
The $ \ q = 0 \ $ case is not mentioned above, as it is "trivial": the zeroes cited become $ \ k =  l =  r = -\frac{q}{p} = 0 \ \ . $
For $ \ p^2 = 4q \ \ , $ we have $ \ k  \ =  \ \frac{-p}{2} \ , \ l \ =  \ \frac{p}{2} \ - \ \frac{\sqrt{8q}}{2} \ , \ $ and $ r \ = \ p \ - \ \sqrt{6q} \ \ . $  A test of the requisite inequalities yields
$$ \frac{-p}{2} \ < \ p \ - \ \sqrt{6q} \ \ [?] \ \ \Rightarrow \ \ \sqrt{6q} \ < \ \frac{3p}{2} \ \ \Rightarrow \ \ 6q \ < \ \frac{9p^2}{4} \  \ \Rightarrow \ \ \frac{24}{9}·q \ < \ p^2 \ = \ 4q \ \ [!] $$
and
$$ p \ - \ \sqrt{6q} \ < \ \frac{p}{2} \ - \ \frac{\sqrt{8q}}{2} \ \ [?] \ \ \Rightarrow \ \ \frac{p}{2} \ < \ \sqrt{6q} \ - \ \sqrt{2q} $$ $$ \Rightarrow \ \ p^2 \ = \ 4q \ = \ 16 · \frac14 · q $$ $$ < \ \ 4 · (\sqrt{6} \ - \ \sqrt{2})^2 · q \ = \ 4 · 2 · (\sqrt{3} \ - \ 1)^2 · q \ = \ 16 · (2 - \sqrt3) · q \ \ [!] \ \ . $$
Thus, $ \ r \ $ lies in $ \ [k \ , \ l] \ $ for this case.
In the general case, we will write $ \ p^2 = \alpha·q \ \Rightarrow \ q = \frac{p^2}{\alpha} \ \ , \ \ \alpha > 4 \ \ , $  making the zeroes under discussion
$$ k  \ =  \ \frac{-p}{2} \ + \ \frac{p·\sqrt{1 \ - \ \frac{4}{\alpha}}}{2} \ , \ l \ =  \ \frac{p}{2} \ - \ \frac{p·\sqrt{1 \ + \ \frac{4}{\alpha}}}{2} \ , \   r \ = \ p \ - \  p·\sqrt{1 \ + \ \frac{2}{\alpha}}  \ \ . $$
Again, checking the inequalities, we find
$$ \frac{p}{2} · \left( -1 \ + \ \sqrt{1 \ - \ \frac{4}{\alpha}} \right) \ < \ p · \left( 1 \ - \ \sqrt{1 \ + \ \frac{2}{\alpha}} \right) \ \ [?] $$
$$  \Rightarrow \ \ -1 \ + \ \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 2 \ - \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ \ \Rightarrow \ \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ + \ \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 3   $$
$$  \Rightarrow \ \ 4 \ + \ \frac{8}{\alpha} \ + \ 1 \ - \ \frac{4}{\alpha} \ + \ 4 · \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 9 $$ $$  \Rightarrow \ \ \frac{1}{\alpha} \ + \ \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 1  \ \ \Rightarrow \ \   \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 1 \ - \ \frac{1}{\alpha} $$
$$ \Rightarrow \  \   1 \ - \ \frac{2}{\alpha}  \ - \  \frac{8}{\alpha^2}  \ < \ 1 \ - \ \frac{2}{\alpha} \ + \ \frac{1}{\alpha^2} \ \ \Rightarrow \ \ - \frac{8}{\alpha^2}  \ < \  \frac{1}{\alpha^2}  $$
and
$$ p · \left( 1 \ - \ \sqrt{1 \ + \ \frac{2}{\alpha}} \right) \ < \ \frac{p}{2} · \left( 1 \ - \ \sqrt{1 \ + \ \frac{4}{\alpha}} \right) \ \ [?] $$
$$  \Rightarrow \ \ 2 \ - \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ < \ 1 \ - \ \sqrt{1 \ + \ \frac{4}{\alpha}} \ \ \Rightarrow \ \ 1 \ < \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ - \ \sqrt{1 \ + \ \frac{4}{\alpha}}   $$
$$  \Rightarrow \ \ 1 \ < \ 4 \ + \ \frac{8}{\alpha} \ + \ 1 \ + \ \frac{4}{\alpha} \ - \ 4 · \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ + \ \frac{4}{\alpha}}  $$
$$ \Rightarrow \ \ \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ + \ \frac{4}{\alpha}} \ < \ 1 \ + \ \frac{3}{\alpha}   $$
$$ \Rightarrow \  \   1 \ + \ \frac{6}{\alpha}  \ + \  \frac{8}{\alpha^2}  \ < \ 1 \ + \ \frac{6}{\alpha} \ + \ \frac{9}{\alpha^2} \ \ \Rightarrow \ \  \frac{8}{\alpha^2}  \ < \  \frac{9}{\alpha^2}  $$
So these inequalities are certainly true for $ \ \alpha > 4 \ \ . $
We may therefore conclude that $ \ k \ < \ r \ < \ l \ \ . $  [It can also be shown by corresponding arguments that this also applies for the other (more widely-separated) roots
$$  k'  \ =  \ \frac{-p}{2} \ - \ \frac{\sqrt{p^2 \ - \ 4q}}{2} \ < \ r \ < \ l' \ =  \ \frac{p}{2} \ + \ \frac{\sqrt{p^2 \ + \ 4q}}{2} \ \ . ] $$
</s>