inkyvoyd
  • inkyvoyd
How does one solve a polynomial absolute value inequality? say, |(x-a)(x-b)(x-c)|>d
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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inkyvoyd
  • inkyvoyd
No joke, I got a prob on a quiz for my AP BC calc class and I miserably got it wrong...
inkyvoyd
  • inkyvoyd
the lesson mentioned inequalities for polynomials, but not polynomial abs value probs.
inkyvoyd
  • inkyvoyd
And, if I'm supposed to use a graphing calculator, tell me, because I honestly have no idea.

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phi
  • phi
I think these are tricky. I would re-write as (x-a)(x-b)(x-c) - d > 0 or -(x-a)(x-b)(x-c) - 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
  • inkyvoyd
Yes, but linear absolute value problems can have extraneous solutions...
phi
  • phi
then check the possible answers in the original equation
inkyvoyd
  • inkyvoyd
So I probs just want a graphing calculator?
phi
  • phi
off hand, I don't know. do you have a specific example?
inkyvoyd
  • inkyvoyd
Let me access my course. Sec
inkyvoyd
  • inkyvoyd
|x^3-x^2+x-1|<3 "Round your answer to three decimal places and place in ascending order with a comma in between." Answer: -0.811
inkyvoyd
  • inkyvoyd
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.
phi
  • phi
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.
inkyvoyd
  • inkyvoyd
Alright. Thanks for looking at it with me.
amistre64
  • amistre64
if you graph the inner polynomial, whenever it goes negative, it just mirrors above the x axis
amistre64
  • amistre64
well, the curve does, then pick a point that is clearly not on the curve to test
amistre64
  • amistre64
|dw:1348175550151:dw|
amistre64
  • amistre64
|dw:1348175643801:dw|
inkyvoyd
  • inkyvoyd
o.o
inkyvoyd
  • inkyvoyd
it's a fake 6 degree polynomial!
inkyvoyd
  • inkyvoyd
|dw:1348176111960:dw|
inkyvoyd
  • inkyvoyd
Mmm.
amistre64
  • amistre64
its just the normal 3 degree with its negative parts popped up; since the absolute value of a number is never negative.
inkyvoyd
  • inkyvoyd
Yeah. and then we draw that line. any way of an analytic solution, or are we just going to have to graph this guy?
amistre64
  • amistre64
im curious why the y=3 line
amistre64
  • amistre64
i see it now
amistre64
  • amistre64
yes, that y=3 line represents where the x axis would be if we subtract 3 from each side
amistre64
  • amistre64
|x^3-x^2+x-1| < 3 consider the values such that (x^3-x^2+x-1) = 3 -(x^3-x^2+x-1) = 3
amistre64
  • amistre64
x^3-x^2+x-4 = 0 and x^3-x^2+x+2 = 0 there is no good way to do a general cubic solution is there
inkyvoyd
  • inkyvoyd
yeah I'm not cardano, I have 30 minutes to do this xD. I think it was a calculator prblem lol
amistre64
  • amistre64
do you need exact or approx ?
inkyvoyd
  • inkyvoyd
approx.
amistre64
  • amistre64
x = 1.743 and -0.811 is what the google gives me as approx zeros
amistre64
  • amistre64
we can test those in the original format
inkyvoyd
  • inkyvoyd
kay
amistre64
  • amistre64
at x=1.743; |x^3-x^2+x-1| -3 = 0.000xxx at x=-0.811; |x^3-x^2+x-1| -3 = 0.00xxxx and of course, when x=0, 1-3 is less than 0 so it should be an interval like: (-0.811,1.743) or however exact you need it to be
inkyvoyd
  • inkyvoyd
I see. So that's what I'm supposed to do. Thank you.
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=%7Cx%5E3-x%5E2%2Bx-1%7C+-3%3C0 it even has the interval as a number line at the bottom
phi
  • phi
Here is what I came up with |x^3-x^2+x-1|<3 we can do this with a graphing calculator Case 1: x^3-x^2+x-1 < 3 x^3-x^2+x-4 < 0 Examine the roots of x^3-x^2+x-4 = 0 Find 1 real root at x= 1.743 Graph of the function shows f(x)<0 for x < 1.743 Case 2: -(x^3-x^2+x-1) < 3 or, negating and flipping the sense of the relational operator x^3-x^2+x-1 > -3 x^3-x^2+x+2 > 0 Find the roots of f(x)= x^3-x^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
  • inkyvoyd
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
phi
  • phi
Sounds like you are ready for these... give 'em hell
inkyvoyd
  • inkyvoyd
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.
inkyvoyd
  • inkyvoyd
It's not even calculus and I got it wrong. xS

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