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

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1the lesson mentioned inequalities for polynomials, but not polynomial abs value probs.

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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1I 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...

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, but linear absolute value problems can have extraneous solutions...

phi
 4 years ago
Best ResponseYou've already chosen the best response.1then check the possible answers in the original equation

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1So I probs just want a graphing calculator?

phi
 4 years ago
Best ResponseYou've already chosen the best response.1off hand, I don't know. do you have a specific example?

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1Let me access my course. Sec

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1x^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)

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, 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.

phi
 4 years ago
Best ResponseYou've already chosen the best response.1I'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.

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1Alright. Thanks for looking at it with me.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0if you graph the inner polynomial, whenever it goes negative, it just mirrors above the x axis

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0well, the curve does, then pick a point that is clearly not on the curve to test

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348175550151:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348175643801:dw

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1it's a fake 6 degree polynomial!

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

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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0im curious why the y=3 line

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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0x^3x^2+x1 < 3 consider the values such that (x^3x^2+x1) = 3 (x^3x^2+x1) = 3

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

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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0do you need exact or approx ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0x = 1.743 and 0.811 is what the google gives me as approx zeros

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0we can test those in the original format

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0at 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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1I see. So that's what I'm supposed to do. Thank you.

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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1Here 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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1I 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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1Sounds like you are ready for these... give 'em hell

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1Too 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.

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1It's not even calculus and I got it wrong. xS
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