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Math Help - solving an inequality for x when an absolute value sign appears on both sides

  1. #1
    Junior Member jonnygill's Avatar
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    solving an inequality for x when an absolute value sign appears on both sides

    the problem is...

    3|x-7|<|x-1|

    i wasn't sure how to solve this problem. after looking in the back of the book (which only shows final answers, not the steps to get there) i determined that they solved by setting |x-7| to -(x-7) and +(x-7) and, in both instances, setting |x-1| to (x-1).

    Why did they not set |x-1| to (-x+1) and (x-1) as well? this is how i approached the problem, which means i worked out four different equations... ++, +-, -+, --

    this is not the correct way to approach the problem, as it results in "no solution."

    I see how they arrived at the solution, i just don't understand why.

    note: oddly, when i changed 3|x-7| to 3x-21 such that..

    3x-21<|x-1| and worked the two equations... 3x-21<x-1 and 3x-21<-x+1 i ended up with a completely different solution set.

    when there is an absolute value sign on both sides of the inequality, how does one know which to "leave alone" aka take the positive value of, and which to substitute in the negative and positive values.
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    Quote Originally Posted by jonnygill View Post
    the problem is... 3|x-7|<|x-1|
    Can you solve 9(x^2-14x+49)<x^2-2x+1~?

    Look at this.
    Last edited by Plato; February 23rd 2011 at 03:45 PM.
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  3. #3
    Junior Member jonnygill's Avatar
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    Quote Originally Posted by Plato View Post
    Can you solve 9(x^2-14x+49)<x^2-2x+1~?
    i don't think i can (right now)

    i tried using the quadratic equation, but i ended up having to take the square root of a negative number and i don't know how to do that yet. i guess it would be in terms of an imaginary number. i suppose this would be an acceptable answer under certain circumstances, but not in this context necessarily.

    also, i don't see the point plato is trying to make. plato's inequality and the original inequality don't appear to be equivalent (as far as i can tell)
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    Quote Originally Posted by jonnygill View Post
    i don't think i can
    Look at this.
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    Junior Member jonnygill's Avatar
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    Quote Originally Posted by Plato View Post
    ok, so that explained the non-real solutions. i would have to study more to understand what is going on. perhaps i will revisit this problem or one like it in the future. for now, i am more interested in knowing why only one absolute value expression was changed to + and - while the other was used only it its positive form.

    i explained more thoroughly what i am trying to ask in the first post of this thread.


    *to be more specific, i will say, how does one solve the inequality for only real solutions*
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    Quote Originally Posted by Plato View Post
    Can you solve 9(x^2-14x+49)<x^2-2x+1~?

    Look at this.
    \displaystyle 9x^2 - 126x + 441 < x^2 - 2x + 1

    \displaystyle 8x^2 - 124x + 440 < 0

    \displaystyle x^2 - \frac{31}{2} + 55 < 0

    \displaystyle x^2 - \frac{31}{2} + \left(-\frac{31}{4}\right)^2 - \left(-\frac{31}{4}\right)^2 + 55 < 0

    \displaystyle \left(x - \frac{31}{4}\right)^2 - \frac{961}{16} + \frac{880}{16} < 0

    \displaystyle \left(x - \frac{31}{4}\right)^2 - \frac{81}{16} < 0

    \displaystyle \left(x - \frac{31}{4}\right)^2 < \frac{81}{16}

    \displaystyle \left|x - \frac{31}{4}\right| < \frac{9}{4}

    \displaystyle -\frac{9}{4} < x - \frac{31}{4} < \frac{9}{4}

    \displaystyle \frac{11}{2} < x < 10.
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  7. #7
    Junior Member jonnygill's Avatar
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    Quote Originally Posted by Prove It View Post
    \displaystyle 9x^2 - 126x + 441 < x^2 - 2x + 1

    \displaystyle 8x^2 - 124x + 440 < 0

    \displaystyle x^2 - \frac{31}{2} + 55 < 0

    \displaystyle x^2 - \frac{31}{2} + \left(-\frac{31}{4}\right)^2 - \left(-\frac{31}{4}\right)^2 + 55 < 0

    \displaystyle \left(x - \frac{31}{4}\right)^2 - \frac{961}{16} + \frac{880}{16} < 0

    \displaystyle \left(x - \frac{31}{4}\right)^2 - \frac{81}{16} < 0

    \displaystyle \left(x - \frac{31}{4}\right)^2 < \frac{81}{16}

    \displaystyle \left|x - \frac{31}{4}\right| < \frac{9}{4}

    \displaystyle -\frac{9}{4} < x - \frac{31}{4} < \frac{9}{4}

    \displaystyle \frac{11}{2} < x < 10.

    thanks, but i don't follow this from beginning to end.
    \displaystyle x^2 - \frac{31}{2} + \left(-\frac{31}{4}\right)^2 - \left(-\frac{31}{4}\right)^2 + 55 < 0
    i have never used this method. i see that you added and subtracted by the same number, which wouldn't change the overall expression. but i don't know why and i don't follow the rest of the solution. how did you know to add and subtract that particular fraction? what's the connection?

    also, i don't see how we got from the original inequality i posted to the inequality that plato posted.

    so all in all, i'd say i'm pretty lost as of right now.

    maybe it would be helpful if i showed how i think it can be solved...

    3|x-7|<|x-1|

    3x-21<x-1 AND -3x+21<x-1

    so we get x<10 AND x>\dfrac{22}{4} OR \dfrac{22}{4}<x<10

    so it seems this method works. but why did we just drop the absolute value sign from |x-1| making it just x-1?

    Why did we do that?

    and why can't we do the reverse? (i.e. drop the absolute value sign from |x-7| and work it out keeping |x-1|)


    note: the book i'm working from wouldn't expect me to solve the equation using prove it's approach. they would expect me to solve it using a simpler approach. that simpler approach is the one i'm after.
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  8. #8
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    It's called completing the square...
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    Junior Member jonnygill's Avatar
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    Quote Originally Posted by Prove It View Post
    It's called completing the square...
    ok, should i take this as "you're doing it wrong... the way i showed you is the only way to solve it" or "this is a potentially more difficult way to solve it, although there is an easier way to solve it that hasn't been discussed. i'm just trying to encourage you to explore more difficult methods of solving the problem."

    i haven't studied completing the square yet, so i don't fully understand how to use it. if that's the only way to solve this problem, it would make my life a teeny-tiny-little bit easier if you just said, "this is the only way to solve it so you better learn how to do it this way if you want to know how to solve the problem."
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    Quote Originally Posted by jonnygill View Post
    ok, should i take this as "you're doing it wrong... the way i showed you is the only way to solve it" or "this is a potentially more difficult way to solve it, although there is an easier way to solve it that hasn't been discussed. i'm just trying to encourage you to explore more difficult methods of solving the problem."

    i haven't studied completing the square yet, so i don't fully understand how to use it. if that's the only way to solve this problem, it would make my life a teeny-tiny-little bit easier if you just said, "this is the only way to solve it so you better learn how to do it this way if you want to know how to solve the problem."
    By no means is it the only way of solving quadratic inequalities, but it is without question the easiest...

    Consider for example

    \displaystyle (x + 3)^2 - 4 < 0.

    You should be able to solve that for \displaystyle x without too much trouble, because the inverse operations are obvious.

    The inverse operations to solve \displaystyle x^2 + 6x + 5 < 0 are not obvious, even though this is the same inequation.

    Completing the square would convert this to the first inequality and thereby make solving the inequation much easier...
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    Quote Originally Posted by jonnygill View Post
    the problem is...

    3|x-7|<|x-1|

    i wasn't sure how to solve this problem. after looking in the back of the book (which only shows final answers, not the steps to get there) i determined that they solved by setting |x-7| to -(x-7) and +(x-7) and, in both instances, setting |x-1| to (x-1).

    Why did they not set |x-1| to (-x+1) and (x-1) as well? this is how i approached the problem, which means i worked out four different equations... ++, +-, -+, --

    this is not the correct way to approach the problem, as it results in "no solution."

    I see how they arrived at the solution, i just don't understand why.

    note: oddly, when i changed 3|x-7| to 3x-21 such that..

    3x-21<|x-1| and worked the two equations... 3x-21<x-1 and 3x-21<-x+1 i ended up with a completely different solution set.

    when there is an absolute value sign on both sides of the inequality, how does one know which to "leave alone" aka take the positive value of, and which to substitute in the negative and positive values.
    You are missing something.

    3|x-7|<|x-1|

    If x>7, both x-7 and x-1 are positive.
    Hence, we evaluate

    3(x-7)<(x-1)

    If x<7, then (x-1) is positive until x reaches 1, however (x-7) is negative

    and so for k<x<7, we definately have

    3|x-7|<|x-1|

    Next, if x<1, both (x-7) and (x-1) are negative, but x-7 is more negative than x-1
    and so |x-7|>|x-1| so 3|x-7|>|x-1| if x<1

    hence 3|x-7| cannot be less than |x-1| if x<1

    therefore we only need examine (x-7)>0 and (x-1)>0
    and (x-7)<0 and (x-1)>0

    Another way to view this is to consider the graphs.

    3(x-7) is linear, has a slope of 3 and crosses the x-axis at x=7.
    3|x-7| has the part of the line 3(x-7) under the x-axis inverted
    so that it is above the x-axis,
    so 3|x-7| is V-shaped with a slope of 3 standing on the x-axis at x=7.

    x-1 is linear with a slope of 1, crossing the x-axis at x=1.
    Hence |x-1| is V-shaped with a slope of 1, standing on the x-axis at x=1.
    Clearly, because of the slopes, these graphs intersect beyond x=1 twice.

    Locating these intersections, which is where 3(x-7)=x-1 and -3(x-7)=x-1
    then in between you have your inequality.
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