inkyvoyd 3 years ago How does one solve a polynomial absolute value inequality? say, |(x-a)(x-b)(x-c)|>d

1. inkyvoyd

No joke, I got a prob on a quiz for my AP BC calc class and I miserably got it wrong...

2. inkyvoyd

the lesson mentioned inequalities for polynomials, but not polynomial abs value probs.

3. inkyvoyd

And, if I'm supposed to use a graphing calculator, tell me, because I honestly have no idea.

4. 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...

5. inkyvoyd

Yes, but linear absolute value problems can have extraneous solutions...

6. phi

then check the possible answers in the original equation

7. inkyvoyd

So I probs just want a graphing calculator?

8. phi

off hand, I don't know. do you have a specific example?

9. inkyvoyd

Let me access my course. Sec

10. 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<x<-0.732 Incorrect Response (-.811, 1.743, -.811,1.743, -0.811,1.743, -0.811, 1.743)

11. 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.

12. 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.

13. inkyvoyd

Alright. Thanks for looking at it with me.

14. amistre64

if you graph the inner polynomial, whenever it goes negative, it just mirrors above the x axis

15. amistre64

well, the curve does, then pick a point that is clearly not on the curve to test

16. amistre64

|dw:1348175550151:dw|

17. amistre64

|dw:1348175643801:dw|

18. inkyvoyd

o.o

19. inkyvoyd

it's a fake 6 degree polynomial!

20. inkyvoyd

|dw:1348176111960:dw|

21. inkyvoyd

Mmm.

22. amistre64

its just the normal 3 degree with its negative parts popped up; since the absolute value of a number is never negative.

23. 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?

24. amistre64

im curious why the y=3 line

25. amistre64

i see it now

26. amistre64

yes, that y=3 line represents where the x axis would be if we subtract 3 from each side

27. 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

28. 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

29. inkyvoyd

yeah I'm not cardano, I have 30 minutes to do this xD. I think it was a calculator prblem lol

30. amistre64

do you need exact or approx ?

31. inkyvoyd

approx.

32. amistre64

x = 1.743 and -0.811 is what the google gives me as approx zeros

33. amistre64

we can test those in the original format

34. inkyvoyd

kay

35. 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

36. inkyvoyd

I see. So that's what I'm supposed to do. Thank you.

37. 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

38. 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

39. 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

40. phi

Sounds like you are ready for these... give 'em hell

41. 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.

42. inkyvoyd

It's not even calculus and I got it wrong. xS