Answer:
To prove: The equation x2+px−1=0 has real and distinct roots for all real values of p.
Consider x2+px−1=0
Discriminant D=p2−4(1)(−1)=p2+4
We know p2≥0 for all values of p
⇒p2+4≥0 (since 4>0)
Therefore D≥0
Hence the equation x2+px−1=0 has real and distinct roots for all real values of p
