x^2 + px + q = 0 has one negative solution and one positive solution. If p = 3, then what are all the possible values of q?

Guest Nov 3, 2021

#1**+1 **

*x^2 + px + q = 0 has one negative solution and one positive solution. If p = 3, then what are all the possible values of q?*

In other words x^{2} +3x + q = 0

(x + 4)(x – 1) p = –4

(x + 5)(x – 2) p = –10

(x + 6)(x – 3) p = –18

(x + 7)(x – 4) p = –28

(x + 8)(x – 5) p = –40

I must not understand the question, because this could go on forever and

you couldn't write down all of them if you keep writing until the end of time.

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Guest Nov 3, 2021

#3**+1 **

*I must not understand the question, because this could go on forever and you couldn't write down all of them if you keep writing until the end of time. *

Hi Ron,

You understand the question correctly. Your answer is correct.

The formal solution is (q < 0)

It’s relationship to (x) is: q = -x (x + 3)

...And most importantly, you remembered to "mind your peas and queues"

GA

--. .-

GingerAle Nov 4, 2021