2 questions here, both kind of inequality related:
1. |1-x|=1-x; I know the answer is x less then or equal to 1, but I don't understand why or how.
2. x^3-6x^2+5x+6 less than or equal to 0. It says to use a sign line to figure out the answers. I know how to get the roots, I just don't know how to get the inequalities.

- anonymous

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- zepdrix

|dw:1441677151594:dw|This is how we define absolute value.

- zepdrix

|dw:1441677243703:dw|

- zepdrix

|dw:1441677300321:dw|So when we have 1-x, it tells us that 1-x is greater than or equal to zero.
Solve for x in this inequality.

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- anonymous

Thanks! I don't understand in the first part though. If |x| = -x when x<0, then wouldn't |x| end up negative, contradicting the | |?

- zepdrix

No. That's a good question though.
\(\large\rm |x|=-\color{orangered}{x}\) when x<0.
So for example:
Let's say that x is some negative value like... -5.
Then,\[\large\rm |-5|=-(\color{orangered}{-5})=5\]See how we NEED the double negative to make it positive?
That's what happens with absolute, you need the OPPOSITE of the negative number.

- zepdrix

You're putting a negative value in for the Orange x, you still have the other negative in front of it

- anonymous

Ahhh I see! They cancel out. Thank you! Let me see if I can apply this to the problem.
What are your thoughts on the second question?

- zepdrix

Hmmm thinking :d

- anonymous

Thanks, let me know what you think!

- zepdrix

Were you able to get the factors from the polynomial? :)

- anonymous

Yep. I got 2, 2+sqrt7, and 2-sqrt7. I did the p/q thing and found the factors of to be 6/1 to be 1,2,3,6, and then synthetically divided.

- zepdrix

Good you found your 2 root,\[\large\rm (x-2)(x^2-4x-3)\le0\]And then quadratic gets you the rest of the way, cool.
So if 2+sqrt7 and 2-sqrt7 are roots, then these must be factors of our polynomial,\[\large\rm (x-2)(x-2+\sqrt7)(x-2-\sqrt7)\le0\]
We have this whole thing being negative when exactly Three or one of the factors is negative.
\(\large\rm (-)(+)(+)\le0\)
\(\large\rm (-)(-)(-)\le0\)
\(\large\rm (-)(-)(+)\cancel{\le}0\)

- zepdrix

A sign line... hmm

- zepdrix

|dw:1441678216003:dw|

- zepdrix

I hope the way I wrote those factors wasn't too confusing :o
It might help us a lil bit here.

- zepdrix

|dw:1441678352725:dw|So when we're less than 2, do you see how that first factor will be negative? :) lol woops

- anonymous

Interesting... I sort of tried this. I just don't know what my answers mean and how they affect the inequality.

- anonymous

But isn't 2 the first factor? Not sure what you mean?

- anonymous

Or do you mean the x-2?

- zepdrix

Yes, x-2 is the first factor of the polynomial.

- anonymous

Ah so X has to be greater than or equal to 2

- anonymous

Ok that makes sense now

- zepdrix

|dw:1441678495872:dw|So our first factor is negative when we're less than 2,
and positive when we're larger than 2.

- zepdrix

|dw:1441678581150:dw|You can do similar with the second factor (which corresponds to 2-sqrt7).
It will be negative when we're smaller than 2-sqrt7, and positive everywhere else.

- anonymous

So we have to see where all these + and - align?

- zepdrix

|dw:1441678686230:dw|Here is our last one, ya?
Negative when smaller than 2+sqrt7,
positive when larger than 2+sqrt7.

- zepdrix

We're negative when we're multiplying 1 or 3 negatives together.
But positive if we're multiplying 2 or 0 negatives together.

- zepdrix

|dw:1441678807351:dw|Something like this, ya? :o

- zepdrix

I'm not sure if this is how they wanted you to do it,
maybe there is a more organized like.. chart method that I'm forgetting about :) lol

- anonymous

I see... thank you. So the inequality would thus state what? How would I express it as an inequality?

- zepdrix

|dw:1441678919118:dw|These are the intervals that satisfy the inequality.
So our inequality is true when: \(\large\rm x\le 2-\sqrt7\)
and

- zepdrix

when \(\large\rm 2\le x\le 2+\sqrt7\)

- anonymous

thank you!

- zepdrix

np! :)
inequalities can be really tricky! :O

- anonymous

They suck!

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