anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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zepdrix
  • zepdrix
|dw:1441677151594:dw|This is how we define absolute value.
zepdrix
  • zepdrix
|dw:1441677243703:dw|
zepdrix
  • 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
  • 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
  • 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
  • zepdrix
You're putting a negative value in for the Orange x, you still have the other negative in front of it
anonymous
  • 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
  • zepdrix
Hmmm thinking :d
anonymous
  • anonymous
Thanks, let me know what you think!
zepdrix
  • zepdrix
Were you able to get the factors from the polynomial? :)
anonymous
  • 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
  • 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
  • zepdrix
A sign line... hmm
zepdrix
  • zepdrix
|dw:1441678216003:dw|
zepdrix
  • zepdrix
I hope the way I wrote those factors wasn't too confusing :o It might help us a lil bit here.
zepdrix
  • zepdrix
|dw:1441678352725:dw|So when we're less than 2, do you see how that first factor will be negative? :) lol woops
anonymous
  • anonymous
Interesting... I sort of tried this. I just don't know what my answers mean and how they affect the inequality.
anonymous
  • anonymous
But isn't 2 the first factor? Not sure what you mean?
anonymous
  • anonymous
Or do you mean the x-2?
zepdrix
  • zepdrix
Yes, x-2 is the first factor of the polynomial.
anonymous
  • anonymous
Ah so X has to be greater than or equal to 2
anonymous
  • anonymous
Ok that makes sense now
zepdrix
  • 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
  • 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
  • anonymous
So we have to see where all these + and - align?
zepdrix
  • zepdrix
|dw:1441678686230:dw|Here is our last one, ya? Negative when smaller than 2+sqrt7, positive when larger than 2+sqrt7.
zepdrix
  • 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
  • zepdrix
|dw:1441678807351:dw|Something like this, ya? :o
zepdrix
  • 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
  • anonymous
I see... thank you. So the inequality would thus state what? How would I express it as an inequality?
zepdrix
  • 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
  • zepdrix
when \(\large\rm 2\le x\le 2+\sqrt7\)
anonymous
  • anonymous
thank you!
zepdrix
  • zepdrix
np! :) inequalities can be really tricky! :O
anonymous
  • anonymous
They suck!

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