discriminant higher maths

Question

question: show that the roots of the equation X²+PX-1=0are always real.

I used the discriminant.

a=1 b=p c=-1

p² - 4*1*-1

P²+4=0

p= -4

P=√4

p=2

am  right because I don't have the answers

Answer 1

x² + px - 1 = 0

 

It is sufficient to show that the discriminant

 

p² - 4(1)(-1) ≥ 0

 

p² - 4(1)(-1) = p² + 4 > 0 for any real p

 

Note the discriminant is positive for any real p, so you will always get 2 distinct real roots for any real p.

 

Answer 2

p^2 +4 is greater than zero for any p belonging to set of R. If the discriminant is greater than zero there are two real solutions.