+0

0
45
3

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?

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         x2 +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.

.

Nov 3, 2021
#2
+1

oops, my eyes must have crossed, I typed a p instead of a q

The above should be

(x + 4)(x – 1)     q = –4

(x + 5)(x – 2)     q = –10

(x + 6)(x – 3)     q = –18

(x + 7)(x – 4)     q = –28

(x + 8)(x – 5)     q = –40

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,

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

--. .-

Nov 4, 2021