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anonymous
 one year ago
2 questions here, both kind of inequality related:
1. 1x=1x; I know the answer is x less then or equal to 1, but I don't understand why or how.
2. x^36x^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
 one year ago
2 questions here, both kind of inequality related: 1. 1x=1x; I know the answer is x less then or equal to 1, but I don't understand why or how. 2. x^36x^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.

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441677151594:dwThis is how we define absolute value.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441677300321:dwSo when we have 1x, it tells us that 1x is greater than or equal to zero. Solve for x in this inequality.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks! 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
 one year ago
Best ResponseYou've already chosen the best response.2No. 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
 one year ago
Best ResponseYou've already chosen the best response.2You're putting a negative value in for the Orange x, you still have the other negative in front of it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks, let me know what you think!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Were you able to get the factors from the polynomial? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep. I got 2, 2+sqrt7, and 2sqrt7. 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
 one year ago
Best ResponseYou've already chosen the best response.2Good you found your 2 root,\[\large\rm (x2)(x^24x3)\le0\]And then quadratic gets you the rest of the way, cool. So if 2+sqrt7 and 2sqrt7 are roots, then these must be factors of our polynomial,\[\large\rm (x2)(x2+\sqrt7)(x2\sqrt7)\le0\] 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
 one year ago
Best ResponseYou've already chosen the best response.2I hope the way I wrote those factors wasn't too confusing :o It might help us a lil bit here.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441678352725:dwSo when we're less than 2, do you see how that first factor will be negative? :) lol woops

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Interesting... I sort of tried this. I just don't know what my answers mean and how they affect the inequality.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But isn't 2 the first factor? Not sure what you mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or do you mean the x2?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes, x2 is the first factor of the polynomial.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah so X has to be greater than or equal to 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok that makes sense now

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441678495872:dwSo our first factor is negative when we're less than 2, and positive when we're larger than 2.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441678581150:dwYou can do similar with the second factor (which corresponds to 2sqrt7). It will be negative when we're smaller than 2sqrt7, and positive everywhere else.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we have to see where all these + and  align?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441678686230:dwHere is our last one, ya? Negative when smaller than 2+sqrt7, positive when larger than 2+sqrt7.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2We're negative when we're multiplying 1 or 3 negatives together. But positive if we're multiplying 2 or 0 negatives together.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441678807351:dwSomething like this, ya? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I'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
 one year ago
Best ResponseYou've already chosen the best response.0I see... thank you. So the inequality would thus state what? How would I express it as an inequality?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441678919118:dwThese are the intervals that satisfy the inequality. So our inequality is true when: \(\large\rm x\le 2\sqrt7\) and

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2when \(\large\rm 2\le x\le 2+\sqrt7\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2np! :) inequalities can be really tricky! :O
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