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.

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) (Show Source): You can put this solution on YOUR website!
From the quadratic formula

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

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

Plug in , ,

Foil and multiply

Combine like terms

Now since we want to have 2 non-real solutions, this means

Set the discriminant less than zero

Subtract 1 from both sides

Combine like terms on the right side

Divide both sides by 4 to isolate p

Reduce

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Answer:
So our answer is (which is approximately in decimal form)

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

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