Write the set of value of 'a' for which the equation

The given quadric equation is $x^{2}+a x-1=0$

Then find the value of a.

Here, $a=1, b=a$ and, $c=-1$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=1, b=a$ and, $c=-1$

$=(a)^{2}-4 \times 1 \times-1$

$=a^{2}+4$

The given equation will have real roots, if $D>0$.

$a^{2}+4>0$

$\Rightarrow a^{2}>-4$ which is true for all real values of $a$.

Therefore, for all real values of $a$, the given equation has real roots.