Determine the values of p for which the quadratic equation

Given:

$2 x^{2}+p x+8=0$

Here,

$a=2, b=p$ and $c=8$

Discriminant $D$ is given by :

$D=\left(b^{2}-4 a c\right)$

$=p^{2}-4 \times 2 \times 8$

$=\left(p^{2}-64\right)$

If $D \geq 0$, the roots of the equation will be real.

$\Rightarrow\left(p^{2}-64\right) \geq 0$

$\Rightarrow(p+8)(p-8) \geq 0$

$\Rightarrow p \geq 8$ and $p \leq-8$

Thus, the roots of the equation are real for $p \geq 8$ and $p \leq-8$.