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Simple maths problem (Greater than/Less than) help

  1. Jan 1, 2009 #1
    x(squared) - 7x - 18 > 0

    Right I know what to do, you factorise the quadratic etc.

    So you get (x + 2)(x - 9) > 0

    I though -2<x<9 but apparently I'm wrong? I always get confused when it comes to using GREATER THAN/ LESS THAN signs and having to find a set of values which satisfy the equation. When are the x values greater than or greater and equal etc. ?

    I'm sorry if I may not be explaining my problem very well, a simple question which commonly occurs in our AS-Level Maths papers but just don't know how to write the set of x values which satisfy the equation, when is it greater than or less than etc.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 1, 2009 #2


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    Draw a picture. (Seriously)

    As x varies, continuous functions cannot change their sign without passing through zero. If there are few zeroes, then that reduces the problem to just a handful of things to check via direct calculation. (and even less if you really know the behavior of the function in question)
  4. Jan 1, 2009 #3


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    For the equation to be greater than zero, what must be true of both of the factors?
  5. Jan 1, 2009 #4


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    You can think of x^2 - 7x - 18 as a function, let's say f(x) = x^2 - 7x - 18. Can you plot the graph of this function, and see where f(x) > 0 holds? This should help.
  6. Jan 1, 2009 #5

    D H

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    Correct (the "apparently I'm wrong" part, that is). Drawing a picture, as Hurkyl and radao suggested, will demonstrate this. Use your graphing calculator, if you have one.

    Both x+2 and x-9 must have the same sign for (x+2)(x-9) to be positive. So, two cases to analyze: x+2 and x-9 are both negative, or x+2 and x-9 are both positive. Can you take it from here?
  7. Jan 1, 2009 #6


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    Another way of looking at it is this: you know that (x+2)(x-9) is equal to 0 when x= -2 or x= 9. Those are the points where (x+2)(x-9) can change sign. Divide the real line into three parts, x< -2, -2< x< 9, and 9< x. Choose one value of x in each interval and check the sign at that point. For example, -3< -2 and (-3)2- 7(-3)- 18= 9+ 21- 18= 12> 0. Since the sign can change only at -2 and 9, x2- 7x - 18> 0 for all x< -2. -2< 0< 9 and (0)2- 7(0)- 18= -18< 0 so x2- 7x- 18< 0 for all x between -2 and 9. Finally, 10> 9 and (10)2- 7(10)- 18= 100- 70- 18= 12> 0. x2- 7x- 18> 0 for all x> 9.
  8. Jan 1, 2009 #7
    I think graph technique is the best but I wasn't allowed to use that in the exam. My professor would use different cases .. solve each of them and in the end, combine everything using AND/OR symbols(a lot of mess and I never really remembered them). So, here's what I used:

    (x+2).(x-9) > 0
    [(x+2) > 0 AND (x-9) > 0] OR [(x+2) < 0 AND (x-9) < 0]
    note the distinction between AND, OR they are boolean
    Now, it's an easier problem to solve. You just need to take care of AND's OR's. Instead of introducing AND/OR in the end, I state them in the beginning.
  9. Jan 3, 2009 #8
    For quadratic functions of x that has solutions to determine which symbols to use depends on the co-efficient of x2.

    A negative co-efficient will show an upside down U-shape and positive will show a U-shape for your curve. So, naturally if it has real roots then f(x) < 0 will be continous for positive co-efficients and broken for negative co-efficients and vica versa. It is a good idea to draw a graph or something because losing marks in an exam for questions like these is really annoying because it is something you can do quite easily.


    x2 -7x - 18
    (x + 2)(x - 9)>0
    Since this is for > 0 it is a broken inequality.
    (x + 2)(x - 9)<0
    Since this is for < 0 it is continous for a positive U shaped curve.

    For negative co-efficents it will be the opposite. Where (-x-2)(x-9)<0


    EDIT: don't try to solve for (x+a)(x+b)>0

    doesn't really work.
    Last edited: Jan 3, 2009
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