Question:
Write the set of value of ' $a$ ' for which the equation $x^{2}+a x-1=0$ has real roots.
Solution:
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.
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