Question 1038912
.
{{{X^2-px+4>0}}} then |p|
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
  {{{x^2 -px + 4}}} = 

= {{{(x-(p/2))^2 - (p/2)^2 + 4}}} = 

= {{{(x-(p/2))^2}}} + {{{(4-(p/2)^2)}}}.   (1)

In (1), the addend  {{{(x-(p/2))^2}}}  can be done an arbitrary small by choosing  "x",  but  the entire sum (1) remains positive, according to the condition.
It means that the second addend  {{{(4-(p/2)^2)}}} > 0,  or,  in other words,
{{{(p/2)^2}}} < 4.

Hense,  {{{abs(p/2)}}} < 2,   or  {{{abs(p)}}} < 4.

QED.

<U>Answer</U>.  |p| < 4.
</pre>

Three plots are shown below with p=3, 4, and 5.

<TABLE> 
  <TR>
  <TD> 

{{{graph( 330, 330, -10.5, 10.5, -10.5, 10.5,
          x^2 -3x + 4, 
          x^2 -4x + 4,
          x^2 -5x + 4
)}}}


<B>Figure</B>. Plots of functions y = {{{x^2 -px + 4}}}, p = 3 (red), 4 (green) and 5 (blue).

  </TD>
  </TR>
</TABLE>