Question

How can i solve this?
2(x-2)(x+1)>0

Answer 1

The original question was:
x(3x+2)>(x+2)^2
so
3(x^2)+2x>(x^2)+4x+4

so:
2x^2 -2x -4 > 0

So :
2(x-2)(x+1)>0

What is the next step?

Answer 2

Either one or the other is bigger than 0, so:
x>2 U x>-1 (???)

Answer 3

No..., actually, x > -1 doesn't do it justice, since
-0.5 > -1
but
if you plug in -0.5, it doesn't work...

Answer 4

I think you distribute 2(x-2) and then you distribute 2(x+1).

Answer 5

Nah... you just have to make sure both factors have the same sign~ that ensures that the resulting product is positive.

Answer 6

how do you solve this problem? I learned inequality but I never learned this.

Answer 7

@osanseviero your steps look right
as terenzreignz said though, the -1 may just be an extraneous answer

Answer 8

Here, let's make a table:
Involving the 'critical values' of the left-part.

Critical values are those values which would make the left-part either equal to zero or undefined.

Thus, the critical values are 2 and -1 as you said.
\[\Large \left.\begin{matrix}&&-\infty&-1&&2&+\infty\\x-2&&-&-&-&0&+\\x+1&&-&0&+&+&+\\(x-2)(x+1)&&+&0&-&0&+\end{matrix}\right.\]

Here's a crude representation...

Answer 9

I have never seen that.

Answer 10

Ok...I knew this method, I wanted to do algebraic

Answer 11

But thanks

Answer 12

Algebraic is a bloody process, and it's not nearly as useful when you have more than two factors, just saying ^_^

Answer 13

so you factor it?

Answer 14

Conveniently, it is already factored j2lie :3

Answer 15

Ok, the table is done. Thanks :)

Answer 16

don't you have to distribute the 2 to each terms in the parentheses?

Answer 17

Nope, it is an extra factor. The answer will be the same

Answer 18

It was actually with the two, I took it out to divide in different components

Answer 19

so the 2 is not needed?

Answer 20

Well, we want the left-part to be positive, but the 2 has no effect on that, since multiplying 2 to anything will not change the sign.

That's why I left the 2 out of the table.

^_^

Answer 21

But j2lie, if you insist...
\[\Large \left.\begin{matrix}&&-\infty&-1&&2&+\infty\\2&&+&+&+&+&+\\x-2&&-&-&-&0&+\\x+1&&-&0&+&+&+\\(x-2)(x+1)&&+&0&-&0&+\end{matrix}\right.\]

Answer 22

I have no idea why you're talking about. I haven't learn this.

Answer 23

Oops...
Anyway, j2lie, just multiply the coloured signs to get the sign at the bottom row, which is what we're looking for:\[\Large \left.\begin{matrix}&&-\infty&-1&&2&+\infty\\2&&\color{blue}+&\color{blue}+&\color{blue}+&\color{blue}+&\color{blue}+\\x-2&&\color{red}-&\color{red}-&\color{red}-&\color{green}0&\color{blue}+\\x+1&&\color{red}-&\color{green}0&\color{blue}+&\color{blue}+&\color{blue}+\\2(x-2)(x+1)&&+&0&-&0&+\end{matrix}\right.\]

Answer 24

what do I do with the numbers?

Answer 25

What numbers? The ones on top?
Those are just markers.

For instance, it shows that 2 is positive everywhere (obviously)
x-2 is negative from negative infinity up to 2, where it is zero. From 2 onwards, it's positive.
x+1 is negative from negative infinity up to -1, where it is zero. From -1 onwards, it's positive.

With these knowledge, we can multiply these three to get the sign of the entire expression
2(x-2)(x+1)

And we want to know at exactly which interval it is positive, which is now shown on the table ^_^

Answer 26

I still don't get it. I'm sorry I haven't learn it yet.

Answer 27

For instance, if we wanted to know the interval where
2(x-2)(x+1) is negative instead, the table shows that it's the interval BETWEEN -1 and 2.

Answer 28

Well don't worry about it, j2lie, just appreciate it ^_^

Answer 29

The interval is positive?

Answer 30

The expression is positive at the intervals from negative infinity to -1, and from 2 to positive infinity.

Answer 31

Yay I got it right.

Answer 32

Thank you @terenzreignz.