Question

Prove that the equation x^2+px-1=0 for every p has two different solutions

Answer 1

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