The set of all real values of $p$ for which the equation $x + 1 = \displaystyle \sqrt{px}$ has exactly one root is

The equation can be written as $x^{ 2 }+(2-p)x+1$

The equation will have exactly 1 root if the equation is a perfect square,

$\Longrightarrow b^{ 2 }-4ac=0$

$\Longrightarrow (2-p)^{ 2 }-4=0$

$\Longrightarrow |2-p|= ±2$

$\Longrightarrow 2-p=2$ =>p=0

or $2-p=-2$ =>p=4

$p=0,4$