# Show that the roots of the equation

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.