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  1. #1
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    inequality question

    How do you solve the inequality x+x>1. I can not remember how to factor it. It doesnt make sense to me how you can have a single x and a negative 1
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  2. #2
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    inequality question

    x+x>1. I can not remmeber how to solve this it doesnt make sense to me how you can have a single x and a negative 1. x(x+1)-1 ???
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  3. #3
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    Quote Originally Posted by msdonnelly View Post
    How do you solve the inequality x+x>1. I can not remember how to factor it. It doesnt make sense to me how you can have a single x and a negative 1
    The inequality can be rearranged to:

    x^2+x-1>0,

    but y=x^2+x-1 is a parabola opening upwards, so y>0 everywhere except between the roots (and at the roots of course)
    of x^2+x-1=0.

    The roots of this quadratic are \frac{-1 \pm \sqrt{5}}{2}, so the inequality is satisfied if x<\frac{-1 - \sqrt{5}}{2} or x>\frac{-1 + \sqrt{5}}{2}

    RonL
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    Quote Originally Posted by msdonnelly View Post
    x+x>1. I can not remmeber how to solve this it doesnt make sense to me how you can have a single x and a negative 1. x(x+1)-1 ???
    Hello,

    this question has been answered twice - and I don't like to work for the waste-bin.

    EB
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by msdonnelly View Post
    x+x>1. I can not remmeber how to solve this it doesnt make sense to me how you can have a single x and a negative 1. x(x+1)-1 ???
    x^2 + x > 1

    x^2 + x - 1 > 0

    (Take a look at the graph I have attached.)

    We are looking for the portion of the x-axis where the graph is above the x-axis.

    We do this algebraically by the following process.

    Solve x^2 + x - 1 = 0
    x = \frac{-1 \pm \sqrt{5}}{2} according to the quadratic formula.

    Now break the x-axis into three parts:
    (- \infty, \frac{-1 - \sqrt{5}}{2} )
    ( \frac{-1 - \sqrt{5}}{2} , \frac{-1 + \sqrt{5}}{2} )
    (\frac{-1 + \sqrt{5}}{2}, \infty )

    We want to test the inequality on each of these intervals. The idea is to pick the easiest number in the interval to do the test on, so for example pick -100000000000000 in the first interval, pick 0 in the second, and pick 100000000000000 in the last interval. (Or any other numbers you like.)
    (- \infty, \frac{-1 - \sqrt{5}}{2} ): x^2 + x - 1 > 0 Check!

    ( \frac{-1 - \sqrt{5}}{2} , \frac{-1 + \sqrt{5}}{2} ): x^2 + x - 1 < 0 Nope!

    (\frac{-1 + \sqrt{5}}{2}, \infty ): x^2 + x - 1 > 0 Check!

    So the solution set is (- \infty, \frac{-1 - \sqrt{5}}{2} ) \cup (\frac{-1 + \sqrt{5}}{2}, \infty ).

    -Dan
    Attached Thumbnails Attached Thumbnails inequality question-parabola.jpg  
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