SOLUTION: for what value(s) of p does the equation x^2+px+1=0 have no real solutions, and two real solutions.

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Question 155038: for what value(s) of p does the equation x^2+px+1=0 have no real solutions, and two real solutions.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
For this discussion, we'll use the discriminant.
For the general quadratic equation,
ax%5E2%2Bbx%2Bc=0
the discriminant is
D=b%5E2-4ac
.
.
If D%3E0 then you have two distinct real roots.
If D=0, you have a double root, one real root occurring twice
If D%3C0, you have two complex roots, that are complex conjugates.
x%5E2%2Bpx%2B1=0
D=p%5E2-4%281%29%281%29
D=p%5E2-4
For no real (complex) solutions,
D%3C0
p%5E2-4%3C0
p%5E2%3C4
Valid interval: (-2%3Cp%3C2)
.
.
.
.
For two real solutions,
D%3E=0
p%5E2%3E=0
p%3E=2 and p%3C=-2
Valid interval:(-infinity,-2]U[2,infinity)