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Quadratics with inequalities from spivak's calc

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Find all numbers x for which

    x2+x+1 > 2 or x2+x-1>0

    2. Relevant equations
    none

    3. The attempt at a solution
    essentially what I did is used the quadratic formula and I got x= [tex]\frac{-1\pm\sqrt{5}}{2}[/tex]

    then I graphed the function and found that x2+x-1 > 0 when x < [tex]\frac{-1-\sqrt{5}}{2}[/tex] or when x > [tex]\frac{-1+\sqrt{5}}{2}[/tex]

    I'm asking if there is a non-graphical method to solving this. It's a simple algebra question I don't remember how to solve.

    This is from Spivak's calculus 3ed, ch 1 problem 4 vi.
     
  2. jcsd
  3. Mar 28, 2010 #2

    Office_Shredder

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    Because quadratic polynomials are continuous functions, they can only change signs when they're zero (this technique works for solving inequalities involving any continuous function)

    So if we call the two roots a and b, a<b, on the intervals

    [tex] (-\infty, a),(a,b),(b,\infty)[/tex]

    The function cannot change sign on any given interval (so on the interval [tex] (-\infty, a)[/tex] it's either always negative or always positive). Then you can just pick a point in each interval and see what the sign is.

    For polynomials specifically, there's another method you can use (this one can be really helpful when determining the sign of a product of a bunch of functions).

    [tex]x^2+x+1=(x-a)(x-b)[/tex].

    The expression (x-a)(x-b) can have its sign determined by considering the sign of each linear factor. x-a is positive if x>a, x-b is positive if x>b. So if x<a (so x<b too), you're multiplying two negative numbers and you get a positive number. If a<x<b, you're multiplying a positive and a negative number, and you get a negative number. If x>b (so x>a also), you're multiplying two positive numbers so get a positive number
     
  4. Mar 28, 2010 #3
    much thanks!
     
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