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Help with Sign Reversals in Solving Quadratic Inequalities

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Here are two, unconnected example problems, having to do with quadratic inequalities. I'm confused b/c I understand how to solve them with an = sign, but as inequalities the book shows extra steps where signs get reversed and a total of ~ 4 possible answers come up instead of 2. I don't see how or why the sign reversal arises and the second set of options comes up.

    I'll go through them sequentially I'm trying to understand the concept, not the particular answer. Please help me understand why the book switches the sign without explaining why! I understand the particular answers but can't derive a rule of thumb. Please help.

    Sorry, using the template doesn't make much sense here, but I'm using it b/c it says to...

    Thanks a lot!

    Example #1) x2>9
    Example #2) 2x2 - 4x - 2 < 4

    2. Relevant equations

    n/a

    3. The attempt at a solution

    Steps the book gives for Example #1 (shown literally)

    i) X2 > 9
    ii) x2 - 9 > 0
    iii) (x+3)(x-3) > 0
    iv) x - 3 < 0 and x + 3 > 0
    -- OR --
    x - 3 > 0 and x + 3 < 0
    ---So---
    x > 3
    or
    x < -3

    My questions: The steps in red are what I don't understand. Why does the book switch the signs and add a second option (the "OR" section)? Why does it not just solve it from step 3 forward, as in:

    (x+3) > 0 ; (x-3) > 0
    x> -3 ; x > 3

    And then cross out x > - 3 b/c it doesn't work, leaving only x > 3?

    How do we know to also add (or look for) x < -3?

    I'm lost.

    Where did this second "or" segment come from and the resulting second answer of x < -3, although I understand how it makes sense plugging it in after the fact.


    Below is example # 2 written out, but my questions are the same - where do the second choices and sign reversals come from and how/when do we know to consider those?

    Thanks and sorry for the long post!!

    ---

    Second Example:

    2x2 - 4x - 2 < 4

    i) 2x2 - 4x - 6 < 0

    ii) 2(x2 - 2x -3) < 0

    iii) 2(x-3)(x+1) < 0

    iv) x - 3 < 0 and x + 1 > 0

    v) x < 3 and x > - 1

    -1 < x < 3
     
  2. jcsd
  3. Oct 14, 2009 #2

    Mark44

    Staff: Mentor

    If the work below is the same as what's in the book, and the problem is as above, the book's answer is wrong. If two factors multiply to make a positive product, then either both factors are positive, or both are negative. They can't have mixed signs as shown below.
    If x + 3 > 0 AND x - 3 > 0, then x > - 3 AND x > 3. If x > -3, it is automatically larger than 3, so that second inequality can be omitted.
    When two factors multiply to make a negative product, then it must be the case that
    1) the first factor is positive AND the second is negative.
    OR
    2) the first factor is negative AND the second is positive.
    If x < 3 and x > -1, then x must be between -1 and 3. IOW -1 < x < 3
     
  4. Oct 14, 2009 #3
    You are right. While the answers are correctly written, I mistyped the signs in the work. It should read:

    x - 3 > 0 and x + 3 > 0

    --or--

    x - 3 < 0 and x + 3 < 0

    --so--

    X > 3
    or
    X<-3

    -------

    Ok, I'm still not getting it, which is frustrating b/c I'm sure it's a fairly simple idea.

    Let me ask it another way.

    In an equation like:

    (x+3)(x-3) = 0 we've got two possibilities [x=-3; x = 3], correct?

    But in an equation like

    (x+3)(x-3) > 0 we've not only got those first two possibilities [[STRIKE]x > -3[/STRIKE]; x > 3]

    but two new possibilities appear when the sign gets switched [x < -3; [STRIKE]x < 3[/STRIKE]], so that the answers are:

    x>3 or x<-3.


    I understand that the two numbers work, but how do I know to flip the sign (to less than in this case) and look for that second set of possibilities? I also don't understand how I know when I'm looking for an "Or" and when I'm looking for an "And" combination?

    Sorry for the remedial questions! Thanks for the explanations!
     
  5. Oct 14, 2009 #4

    ideasrule

    User Avatar
    Homework Helper

    For (x+3)(x-3) > 0 (or indeed any inequality), you need to ask yourself what the signs of the terms have to be to satisfy the inequality. If (x+3)(x-3) is to be positive, then either both terms have to be positive or both have to be negative; if one is positive and the other negative, (x+3)(x-3) would turn out negative. So if x+3 is negative, x has to be smaller than -3; if x-3 is negative, x has to be smaller than 3. So for both to be negative, x < -3. For x+3 to be positive, x has to be greater than -3; for x-3 to be positive, x must be greater than 3. So x > 3.

    The solution to the inequality is x<-3 and x>3; if either is true, (x+3)(x-3) turns out positive. Even though I spent a whole paragraph reasoning this out, I find this reasoning quick and easy. If you don't, you might prefer to think about it in this way: y=(x+3)(x-3) is a parabola that faces up, with roots -3 and +3. it's above the y-axis when x<-3 or x>3, so that's the solution to the inequality.
     
  6. Oct 14, 2009 #5

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm going to answer this question alone, because it's a point people sometimes have trouble getting.

    In the previous step (ignoring their sign mistake), the problem was broken into two cases.

    Once you break a problem into two cases, you solve each case separately as if it was its own problem. In this case (after fixing the sign mistake), you have:

    Example problem #1a: Solve the system of equations
    x² > 9 and x+3 > 0 and x-3 > 0​
    Example problem #1b: Solve the system of equations
    x² > 9 and x+3 < 0 and x-3 < 0​

    Then the solution set for example problem #1 consists of everything in the solution set to problem #1a together with everything in the solution set to problem #1b.

    This comes up all the time. You've even done it before: how would you solve the following problem:
    Find all solutions to x² - 9 = 0​
    ? If you started by factoring:
    If x²-9 = 0, then (x-3)(x+3) = 0​
    then you would probably continue to split it into two cases:
    If (x-3)(x+3) = 0, then x-3=0 or x+3=0​
    and then you would solve each case separately
    If x-3=0 then x=3.
    If x+3=0 then x=-3.​
    The combine:
    If x²-9 = 0, then x=3 or x=-3​
    Then you would plug both in to see if they are solutions
    If x=3 then x²-9 = 3²-9 = 0.
    If x=-3 then x²-9 = (-3)²-9 = 0​
    join the two cases:
    If (x=3 or x=-3) then x²-9 = 0.​
    and then you would finally report:
    (x²-9 = 0) if and only if (x=3 or x=-3).​
     
  7. Oct 14, 2009 #6

    Mark44

    Staff: Mentor

    There's a big difference between equations and inequalities. In a quadratic equation you can have at most two solutions. In a quadratic inequality like you're working with, there are generally an infinite number of solutions.

    Let's take one of your examples first.
    (x + 3)(x - 3) > 0

    For this inequality (it's not an equation) to be satisfied it must be true that:
    {Both factors are positive.} OR {Both factors are negative.}

    The compound statement above can be translated to the following compound statement:
    {The first factor is positive AND the second factor is positive.}
    OR
    {The first factor is negative AND the second factor is negative.}

    In mathematical terms, this is
    {x + 3 > 0 AND x - 3 > 0} OR {x + 3 < 0 AND x - 3 < 0}

    An equivalent way to say this is
    {x > -3 AND x > 3} OR {x < -3 AND x < 3}

    Looking at the one on the left, every number that's bigger than 3 is automatically bigger than -3, so we can omit the x > -3 inequality.
    As an example, suppose your favorite uncle said he was going to give you some money as a gift, and wants you to guess how much. If you say "Is it more than $20?" and he says yes, you wouldn't both to ask later if the gift is $10, would you? If it's more than $20, it is automatically more than $10. That's all that's going on when I say we can ignore the x > -3 inequality. There is no rule about flipping signs or whatever involve here.

    Similarly, for the one on the right, if x < -3, we know for a fact that it's also less than 3, so the x < 3 inequality can be omitted.

    The longer symbolic statement above can be reduced to this:
    x > 3 OR x < -3
    If you draw this set of numbers on the number line, you'll have the ones filled in to the left of -3 together with the ones that are to the right of 3, with none filled in between -3 and 3.

    For a second example, let's look at (x - 2)(x + 1) < 0
    I'm hopeful that you understood my explanation for the previous example, so I'll go a bit faster through this one.

    For the product in this example to be negative, the two factors have to be of different signs.
    {First factor has to be positive AND Second factor has to be negative}
    OR
    {First factor has to be negative AND Second factor has to be positive}

    In symbolic form:
    {x - 2 > 0 AND x + 1 < 0} OR {x - 2 < 0 AND x + 1 > 0}
    Simplifying:
    {x > 2 AND x < - 1} OR {x < 2 AND x > -1}
    The one on the left is impossible. There are no numbers that are simultaneously larger than 2 and less than -2
    The one on the right is possible as long as x is less than 2 and is larger than -1. This happens precisely for those numbers that are between -1 and 2; i.e., for -1 < x < 2.

    The upshot is that you always need to look for all possibilities (that's where the ORs come in in my work) and there is no inequality sign flipping. All of this is very logical, so if you can reason through the possibilities, you can do it. The ANDS come because I'm specifying that BOTH factors (this one AND that one) are positive or negative or whatever). The ORs come because there might be more than one way to get a product of a certain sign.

    The technique I've described can also be extended to inequalities that involve three factors. The only real difference is that you have more possibilities of having three factors be positive or negative.
    Hope that helps.
     
  8. Oct 14, 2009 #7
    Great! These last responses are straightforward and make a lot of sense to me. Thanks a lot to all of you!
     
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