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Math Help - Inequalities

  1. #1
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    Inequalities

    Hi,

    I am extremely confused about inequalities and would appreciate it if someone could help me out. I am not sure how to even phrase what is confusing me so I'll throw out a few example questions:

    Code:
    x^2 - 5x - 14 > 0
    I suspect that for the 2nd and 3rd terms the factors would be (x+7) and (x-2), but I do not know what to do with the first term.

    Another example:

    Code:
    (3x+2)(x-3) < 0
    In this case I have determined the following:

    Code:
    x < -2/3 and x < 3
    and
    x > -2/3 and x > 3
    I do not know why I need those or what they mean, but my textbook said that I need to have both a positive and negative case, thus flipping the direction of the inequality sign. According to WolframAlpha's inequality solver the answer is: -2/3 < x < 3. I can sort of see how they determined the answer from my own findings, but what I do not understand is HOW or WHY they knew to choose x < 3 and not x > 3, and x > -2/3 instead of x < -2/3.

    Thanks.
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  2. #2
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    Re: Inequalities

    how are things out on Jupiter? Chilly I imagine.

    What you've first for your first question doesn't make too much sense though you did identify the correct factorization.

    $x^2-5x-14=(x-7)(x+2)>0$

    The key to this is nothing that in order for the product to be greater than 0 either both factors are greater than 0, or both factors are less than 0.

    so $\left\{ (x-7)>0 \wedge (x+2)>0\right \} \vee \left\{ (x-7)<0 \wedge (x+2)<0\right \}$ ($\vee$ is OR, $\wedge$ is AND.)

    $\left \{(x>7) \wedge (x>-2) \right \} \vee \left \{(x<7) \wedge (x<-2) \right \}$

    $(x>7) \vee (x<-2)$

    The second problem you solve in the same way except that in this case the factors have to be of opposite signs the make the product less than 0. So you have to consider the two cases of

    factor 1 positive, factor 2 negative
    factor 2 negative, factor 2 positive
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  3. #3
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    Re: Inequalities

    Quote Originally Posted by romsek View Post
    how are things out on Jupiter? Chilly I imagine.

    What you've first for your first question doesn't make too much sense though you did identify the correct factorization.

    $x^2-5x-14=(x-7)(x+2)>0$

    The key to this is nothing that in order for the product to be greater than 0 either both factors are greater than 0, or both factors are less than 0.

    so $\left\{ (x-7)>0 \wedge (x+2)>0\right \} \vee \left\{ (x-7)<0 \wedge (x+2)<0\right \}$ ($\vee$ is OR, $\wedge$ is AND.)

    $\left \{(x>7) \wedge (x>-2) \right \} \vee \left \{(x<7) \wedge (x<-2) \right \}$

    $(x>7) \vee (x<-2)$

    The second problem you solve in the same way except that in this case the factors have to be of opposite signs the make the product less than 0. So you have to consider the two cases of

    factor 1 positive, factor 2 negative
    factor 2 negative, factor 2 positive
    Hey thanks for the response. I have a few questions about the points you brought up. For the first question, why is the answer (x > 7) OR (x < -2)? Why couldn't the answer also be (x < 7) OR (x > -2)? If they both have to be greater or both have to be less than zero why is there one taken from each? i.e.: {(x>7)∧(x>−2)}∨{(x<7)∧(x<−2)}

    For the second question, why do they need to be opposite signs to make the product less than 0? Would this not make it true: (x < -2/3 and x < 3), since -2/3 and 3 are of opposite signs?
    Last edited by VinH; March 17th 2014 at 10:23 AM.
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  4. #4
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    Re: Inequalities

    Quote Originally Posted by VinH View Post
    I have a few questions about the points you brought up. For the first question, why is the answer (x > 7) OR (x < -2)? Why couldn't the answer also be (x < 7) OR (x > -2)? If they both have to be greater or both have to be less than zero why is there one taken from each? i.e.: {(x>7)∧(x>−2)}∨{(x<7)∧(x<−2)}
    Have a look at this.
    Thanks from VinH
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  5. #5
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    Re: Inequalities

    Quote Originally Posted by VinH View Post
    Hey thanks for the response. I have a few questions about the points you brought up. For the first question, why is the answer (x > 7) OR (x < -2)? Why couldn't the answer also be (x < 7) OR (x > -2)? If they both have to be greater or both have to be less than zero why is there one taken from each? i.e.: {(x>7)∧(x>−2)}∨{(x<7)∧(x<−2)}

    For the second question, why do they need to be opposite signs to make the product less than 0? Would this not make it true: (x < -2/3 and x < 3), since -2/3 and 3 are of opposite signs?
    to make the product greater than zero either both factors are positive OR both factors are negative. Either one works.

    If x>7, then x is automatically >-2. Hopefully that is obvious.

    Similarly if x<-2, then x<7.

    for the second question if a product of two numbers is less than 0, i.e. negative, then 1 and only 1 of the factors must be negative. It doesn't matter which but only one of them. So either the first factor is negative and the second positive, or the first is positive and the second negative. Either will produce a negative product. This is pretty basic stuff.
    Thanks from VinH
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  6. #6
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    Re: Inequalities

    Quote Originally Posted by VinH View Post
    Hey thanks for the response. I have a few questions about the points you brought up. For the first question, why is the answer (x > 7) OR (x < -2)? Why couldn't the answer also be (x < 7) OR (x > -2)?
    Well, first, ALL numbers satisfy "(x< 7) or (x> -2)" since every number that is NOT less than 7 is larger than -2.

    If they both have to be greater or both have to be less than zero
    If x is greater than 7 then it is also greater than -2. If x is less than -2 then it is also less than 7.

    why is there one taken from each? i.e.: {(x>7)∧(x>−2)}∨{(x<7)∧(x<−2)}

    For the second question, why do they need to be opposite signs to make the product less than 0?
    You should have learned in basic arithmetic: the product of any two positive numbers is positive and the product of any two negative numbers is also positive. In order that the product be negative, the two numbers must be of opposite sign.

    Would this not make it true: (x < -2/3 and x < 3), since -2/3 and 3 are of opposite signs?
    You seem not to understand what "inequality" means. The numbers -3 and -4 satisfy x< -2/3 and x< 3 but have the same signs. The "endpoints", -2/3, and 3, are not the only numbers you need to look at. In fact, "x< -2/3 and x< 3" is exactly the same as "x< 3" by itself. If you do not understand that, you do not understand what "<" means.
    Last edited by HallsofIvy; March 17th 2014 at 11:12 AM.
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  7. #7
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    Re: Inequalities

    Quote Originally Posted by HallsofIvy View Post
    Well, first, ALL numbers satisfy "(x< 7) or (x> -2)" since every number that is NOT less than 7 is larger than -2.


    If x is greater than 7 then it is also greater than -2. If x is less than -2 then it is also less than 7.


    You should have learned in basic arithmetic: the product of any two positive numbers is positive and the product of any two negative numbers is also positive. In order that the product be negative, the two numbers must be of opposite sign.


    You seem not to understand what "inequality" means. The numbers -3 and -4 satisfy x< -2/3 and x< 3 but have the same signs. The "endpoints", -2/3, and 3, are not the only numbers you need to look at. In fact, "x< -2/3 and x< 3" is exactly the same as "x< 3" by itself. If you do not understand that, you do not understand what "<" means.
    I understand what inequality means, I just wasn't getting the logic behind romsek's statement. I don't know "why", it's just one of those things that happen from time to time, especially when tired. Basic multiplication rules are not beyond me, don't worry. As far as the "endpoints" thing goes, that was actually the first thing I looked at for other questions and in those it made sense.

    Quote Originally Posted by romsek View Post
    to make the product greater than zero either both factors are positive OR both factors are negative. Either one works.

    If x>7, then x is automatically >-2. Hopefully that is obvious.

    Similarly if x<-2, then x<7.

    for the second question if a product of two numbers is less than 0, i.e. negative, then 1 and only 1 of the factors must be negative. It doesn't matter which but only one of them. So either the first factor is negative and the second positive, or the first is positive and the second negative. Either will produce a negative product. This is pretty basic stuff.
    I suppose difficulty is relative. My biggest issue is that I am not always good at extracting the meaning behind a formula, and while I understand the base concepts I will often overcomplicate things. That, and being terrible at math in general make for some confusing situations. Anyway, thank you for the input. Clearly I have a lot of improving to do.
    Last edited by VinH; March 17th 2014 at 11:22 AM.
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  8. #8
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    Re: Inequalities

    Hi Vin

    I like to solve inequations as follows.

    First, I solve the associated equation. So in this case I would start with

    $x^2 - 5x - 14 = 0 \implies (x - 7)(x + 2) = 0 \implies x = 7\ or\ x = -\ 2.$ No sweat, right?

    So I know that 7 and - 2 are NOT solutions of the inequation $x^2 - 5x - 14 > 0.$

    Now that leaves three sub-domains to consider, x < - 2, - 2 < x < 7, and x > 7.

    If x < - 2, then $x^2 - 5x - 14 = (x - 7)(x + 2)$ is the product of two negative terms and so is positive. Simple, right?

    If - 2 < x < 7, then $x^2 - 5x - 14 = (x - 7)(x + 2)$ is the product of a negative term (x - 7) and a positive term (x + 2) and so is negative. Again, simple.

    If x > 7, then $x^2 - 5x - 14 = (x - 7)(x + 2)$ is the product of two positive terms and so is positive.. Again, simple.

    Now this is exactly what Romsek did. But it might be helpful to think in terms of a general procedure.

    Solve the associated equation. Those solutions will NOT satisfy the inequation, but they will divide the number line up into sub-domains. Evaluate each sub-domain to determine whether it does or does not satisfy the inequation.
    Thanks from VinH
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