Question

hi:) I’m not sure if this is correct. For the justifying part , must I still use the discriminant or can I just say that since the coefficient of P is positive , b^2 - 4ac > 0 ?

Answer 1

You did it right (almost, I got 21 instead of 19) but didn't finish.  You need to show your discriminant is never negative.

x² + (p+1)x = 5-2p

x² + (p+1)x +(2p-5) =0

Real roots mean a positive (or at least non-negative) discriminant:

D = b² - 4ac = (p+1)² - 4(1)(2p - 5) = p² + 2p + 1 - 8p + 20

D =  p² - 6p + 21

It's not totally obvious that D>0; we prove that by completing the square by noting

(p-3)² = p² - 6p + 9

so

p² - 6p = (p-3)² - 9.

D =  p² - 6p + 21

D =  (p-3)² - 9 + 21

D = (p-3)² + 12

Now we clearly see D>0 always because the squared term can't be negative, so D is always at least 12.   We always get two distinct real roots.