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Math Help - What could x be

  1. #1
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    What could x be

    What could x be in here?
    x(x-6) < -6

    Excuse me for my bad explanation, my English is bad.
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  2. #2
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    Quote Originally Posted by klik11 View Post
    What could x be in here?
    x(x-6) < -6

    Excuse me for my bad explanation, my English is bad.
    x^2 - 6x + 6 < 0

    Draw a graph of y = x^2 - 6x + 6 and get it's x-intercepts. You want the values of x for which y < 0 (that is, the parabola is below the x-axis).
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  3. #3
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    Quote Originally Posted by klik11 View Post
    What could x be in here?
    x(x-6) < -6

    Excuse me for my bad explanation, my English is bad.
    x^2 - 6x < -6

    x^2 - 6x + 9 < -6 + 9

    (x - 3)^2 < 3

    |x-3| < \sqrt{3}

    -\sqrt{3} < x-3 < \sqrt{3}

    3-\sqrt{3} < x < 3 + \sqrt{3}
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  4. #4
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    Thank you both
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  5. #5
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    Yet another way: x(x- 6)< -6 is the same as x^2- 6x+ 6< 0 which is the same as x^2- 6x+ 6= x^2- 6x+ 9- 3= (x- 3)^2- 3= (x-3)^2- \sqrt{3}^2 = (x- 3- \sqrt{3})(x- 3+ \sqrt{3})<  0 (I used the fact that [tex]a^2- b^2= (a- b)(a+ b)).

    The product of two numbers is negative if and only if one is positive and the other negative. Since the two factors are "x- 3" plus and minus the same thing, obviously x- 3+ \sqrt{3}> x- 2- \sqrt{3}. The first must be positive and the second negative: we must have x- 3- \sqrt{3}< 0 and x- 3+ \sqrt{3}> 0.

    From the first x< 3+ \sqrt{3} and from the second x> 3- \sqrt{3}. Together, those give, as before, 3- \sqrt{3}< x< 3+ \sqrt{3}.

    klik11, you English is excellent. Certainly compared with my (Put whatever language you want here)!
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