SOLUTION: For what value(s) of 'p' would the equation px^2 + (2p+1)x+p=0 have two non-real roots? I think it is p=-1/4 but I'm only guessing.

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Question 119096: For what value(s) of 'p' would the equation px^2 + (2p+1)x+p=0 have two non-real roots? I think it is p=-1/4 but I'm only guessing.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
From the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

the discriminant consists of all of the terms in the square root. So the discriminant is

D=b%5E2-4ac

the discriminant tells us how many solutions (and what type of solutions) we can expect for any quadratic.


So let's find the discriminant for px%5E2+%2B+%282p%2B1%29x%2Bp


D=%282p%2B1%29%5E2-4%2Ap%2Ap Plug in a=p, b=2p%2B1, c=p

D=4p%5E2%2B4p%2B1-4p%5E2 Foil and multiply

D=4p%2B1 Combine like terms



Now since we want to have 2 non-real solutions, this means D%3C0


3p%2B1%3C0 Set the discriminant less than zero



4p%3C0-1Subtract 1 from both sides


4p%3C-1 Combine like terms on the right side


p%3C%28-1%29%2F%284%29 Divide both sides by 4 to isolate p



p%3C-1%2F4 Reduce

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Answer:
So our answer is p%3C-1%2F4 (which is approximately p%3C-0.25 in decimal form)



So you're on the right track, but you have to remember that values such as p=-10 work also since -10 is less than -1%2F4