Question

Prove that the equation $x^2+px-1=0$ has real and distinct roots for all real values of $p$.

Answer 1

To prove: The equation $x^2+px-1=0$ has real and distinct roots for all real values of $p$.

Consider $x^2+px-1=0$

Discriminant $D=p^2-4\left(1\right)(-1)=p^2+4$

We know $p^2\ge 0$ for all values of $p$

$\Rightarrow p^2+4\ge 0$ (since $4>0$)

Therefore $D\ge 0$

Hence the equation $x^2+px-1=0$ has real and distinct roots for all real values of $p$.