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inkyvoyd
Group Title
How does one solve a polynomial absolute value inequality? say,
(xa)(xb)(xc)>d
 one year ago
 one year ago
inkyvoyd Group Title
How does one solve a polynomial absolute value inequality? say, (xa)(xb)(xc)>d
 one year ago
 one year ago

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inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
No joke, I got a prob on a quiz for my AP BC calc class and I miserably got it wrong...
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
the lesson mentioned inequalities for polynomials, but not polynomial abs value probs.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
And, if I'm supposed to use a graphing calculator, tell me, because I honestly have no idea.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I think these are tricky. I would rewrite as (xa)(xb)(xc)  d > 0 or (xa)(xb)(xc)  d > 0 multiply out and then factor to find the zeros then look for intervals where the product of the factors are positive very messy...
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Yes, but linear absolute value problems can have extraneous solutions...
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
then check the possible answers in the original equation
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
So I probs just want a graphing calculator?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
off hand, I don't know. do you have a specific example?
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Let me access my course. Sec
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
x^3x^2+x1<3 "Round your answer to three decimal places and place in ascending order with a comma in between." Answer: 0.811<x<0.732 Incorrect Response (.811, 1.743, .811,1.743, 0.811,1.743, 0.811, 1.743)
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
yeah, and my answer was from wolfram. I have no idea what I was supposed to do. and the lesson was not helpful at all. Didn't even mention it.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I'll think about it, as this is not something I've looked at very carefully. But I not that x=1 gives 0 so 1 is part of the solution space. Maybe by tomorrow I'll have something to post.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Alright. Thanks for looking at it with me.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
if you graph the inner polynomial, whenever it goes negative, it just mirrors above the x axis
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
well, the curve does, then pick a point that is clearly not on the curve to test
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348175550151:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348175643801:dw
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
it's a fake 6 degree polynomial!
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
dw:1348176111960:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
its just the normal 3 degree with its negative parts popped up; since the absolute value of a number is never negative.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Yeah. and then we draw that line. any way of an analytic solution, or are we just going to have to graph this guy?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
im curious why the y=3 line
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i see it now
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
yes, that y=3 line represents where the x axis would be if we subtract 3 from each side
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
x^3x^2+x1 < 3 consider the values such that (x^3x^2+x1) = 3 (x^3x^2+x1) = 3
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
x^3x^2+x4 = 0 and x^3x^2+x+2 = 0 there is no good way to do a general cubic solution is there
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
yeah I'm not cardano, I have 30 minutes to do this xD. I think it was a calculator prblem lol
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
do you need exact or approx ?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
x = 1.743 and 0.811 is what the google gives me as approx zeros
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
we can test those in the original format
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
at x=1.743; x^3x^2+x1 3 = 0.000xxx at x=0.811; x^3x^2+x1 3 = 0.00xxxx and of course, when x=0, 13 is less than 0 so it should be an interval like: (0.811,1.743) or however exact you need it to be
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
I see. So that's what I'm supposed to do. Thank you.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=%7Cx%5E3x%5E2%2Bx1%7C+3%3C0 it even has the interval as a number line at the bottom
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Here is what I came up with x^3x^2+x1<3 we can do this with a graphing calculator Case 1: x^3x^2+x1 < 3 x^3x^2+x4 < 0 Examine the roots of x^3x^2+x4 = 0 Find 1 real root at x= 1.743 Graph of the function shows f(x)<0 for x < 1.743 Case 2: (x^3x^2+x1) < 3 or, negating and flipping the sense of the relational operator x^3x^2+x1 > 3 x^3x^2+x+2 > 0 Find the roots of f(x)= x^3x^2+x+2 =0 1 real root at x= 0.811 Graph shows f(x) > 0 for x > 0.811 Intersection of both regions x < 1.743 x > 0.811 Gives 0.811 < x < 1.743
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
I get why wolfram gave me the wrong answer. Cause it split it up into two inequalities. So mad lol. and @phi , thanks for the answer :D
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Sounds like you are ready for these... give 'em hell
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Too late :o I got into BC calc 2.5 weeks late and I'm doing catch up work. that right there was week 1 or something, and I'm at week 2.5 tomorrow ;) But thankfully I understand the concept, so I can take it on in event of unit test.
 one year ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
It's not even calculus and I got it wrong. xS
 one year ago
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