Question

(x+4)+2/(x+1) < 0,solve for x .
very very urgent
@terenzreignz

Answer 1

pls help me

Answer 2

Is it this:
\[\Large (x+4)+\frac{2}{x+1}< 0 \]?

Answer 3

yes

Answer 4

Well, first, you need to combine them into one fraction.

Answer 5

listen its coming (x+3) (x+2) (x+1)<0,then how to find the range of x

Answer 6

What's coming?
Anyway, that's rather simple, you want the product
(x+3)(x+2)(x+1) to be negative so you need either one or three of these factors to be negative.

Answer 7

listen the answer given is -2<x<-1 and -3<x,how?????????????

Answer 8

lol okay, the first step is to find the critical numbers of
(x+3)(x+2)(x+1)
Critical numbers are values of x which would make that product zero.

Answer 9

x=-1,-2 -3

Answer 10

That's right :)
Now, let's make a table
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&&&&&&&\\x+2&&&&&&&\\x+1&&&&&&&\\(x+3)(x+2)(x+1)\end{matrix}\right.\]

Answer 11

OKay? Now, what is the critical number associated with x+3?

Answer 12

infinity

Answer 13

No, I mean, when is x + 3 equal to zero, when x = -3, right?

Answer 14

yap

Answer 15

Okay, so it's zero when x = -3
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&&\color{red}0&&&&&\\x+2&&&&&&&\\x+1&&&&&&&\\(x+3)(x+2)(x+1)\end{matrix}\right.\]

Answer 16

yes

Answer 17

So, before -3, what values does x+3 take?
Positive or negative

Answer 18

negative

Answer 19

that's right :)
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&&&&&&&\\x+1&&&&&&&\\(x+3)(x+2)(x+1)\end{matrix}\right.\]
And positive after -3.

Next, what's the critical number for x+2?

Answer 20

0

Answer 21

No, I mean, when is x+2 = 0 ?

Answer 22

x=-2

Answer 23

Right. So it's zero here
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&&&&\color{red}0&&&\\x+1&&&&&&&\\(x+3)(x+2)(x+1)\end{matrix}\right.\]
To the left, is it positive or negative?

Answer 24

positive
but i am still confused

Answer 25

No... I mean, for values less than -2, then x + 2 would be negative, right?
Try -3
-3 + 2 = -1 which is negative.
Try -2.5
-2.5 + 2 = -0.5 which is negative.

It's negative for all values of x less than -2

Answer 26

\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&-&-&-&\color{red}0&+&+&+\\x+1&&&&&&&\\(x+3)(x+2)(x+1)\end{matrix}\right.\]

Answer 27

-2<x<-1 and -3<x,how?????????????

Answer 28

Yes, we're getting there... :)
Now just do the last row, when is x+1 equal to zero?

Answer 29

x=-1

Answer 30

Right, and it's negative for all values of x less than -1, etc
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&-&-&-&\color{red}0&+&+&+\\x+1&-&-&-&-&-&\color{red}0&+\\(x+3)(x+2)(x+1)\end{matrix}\right.\]

Answer 31

so

Answer 32

Well, over here, we simply have the product of the three, in this row:
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&-&-&-&\color{red}0&+&+&+\\x+1&-&-&-&-&-&\color{red}0&+\\(x+3)(x+2)(x+1)&?&?&?&?&?&?&?\end{matrix}\right.\]

Answer 33

then

Answer 34

We just multiply signs LOL
Over here in the first column
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&-&-&-&\color{red}0&+&+&+\\x+1&-&-&-&-&-&\color{red}0&+\\(x+3)(x+2)(x+1)&\color{blue}?&?&?&?&?&?&?\end{matrix}\right.\]
We have three negative signs... what's negative times negative times negative?

Answer 35

thanx

Answer 36

We're actually not yet done, but if you think you can take it from here, then, be my guest :P

Answer 37

no,explain it to me

Answer 38

First, what is negative x negative x negative?
Is it positive or negative?

Answer 39

-

Answer 40

That's right.
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&-&-&-&\color{red}0&+&+&+\\x+1&-&-&-&-&-&\color{red}0&+\\(x+3)(x+2)(x+1)&-&?&?&?&?&?&?\end{matrix}\right.\]

Answer 41

Now, here, it's zero, because one of the factors is zero:
\[\Large \left.\begin{matrix}&-\infty&-3&&-2&&-1&\infty\\x+3&-&\color{red}0&+&+&+&+&+\\x+2&-&-&-&\color{red}0&+&+&+\\x+1&-&-&-&-&-&\color{red}0&+\\(x+3)(x+2)(x+1)&-&\color{red}0&\color{blue}?&?&?&?&?\end{matrix}\right.\]

What about there^?
(third column)

Answer 42

ok i got it.bye

Answer 43

Oh... okay then :)