Question:
Show that the roots of the equation $x^{2}+p x-q^{2}=0$ are real for all real value of $p$ and $q$.
Solution:
Given:
$x^{2}+p x-q^{2}=0$
Here,
$a=1, b=p$ and $c=-q^{2}$
Discriminant $D$ is given by :
$D=\left(b^{2}-4 a c\right)$
$=p^{2}-4 \times 1 \times\left(-q^{2}\right)$
$=\left(p^{2}+4 q^{2}\right)>0$
$D>0$ for all real values of $p$ and $q$.
Thus, the roots of the equation are real.