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.
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