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Math Help - [SOLVED] Solution Set

  1. #1
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    [SOLVED] Solution Set

    Find the solution set of x + \frac{1}{4x} < 1. The answer they gave was {x : x < 0}
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  2. #2
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    Assume x>0
    Then by AM-GM inequality,
    x+\frac{1}{4x}\ge 2*\sqrt{x*\frac{1}{4x}}
    x+\frac{1}{4x}\ge 1

    Hence we obtain a contradiction.

    So, x<0.
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  3. #3
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    Quote Originally Posted by mark1950 View Post
    Find the solution set of x + \frac{1}{4x} < 1. The answer they gave was {x : x < 0}
    x+\frac{1}{4x}-1<0

    \frac{4x^2}{4x}+\frac{1}{4x}-\frac{4x}{4x}<0

    \frac{4x^2-4x+1}{4x}<0

     <br />
\frac{(2x-1)^2}{4x}<0<br />

    Note that this (2x-1)^2 is always positive so now lets look at the denominator (4x) so the only possible solution here is x<0
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  4. #4
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    Wow. thanks. How about if the numerator was (2x - 1)^3 as in always negative? What would the new solution set be? Thanks.
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  5. #5
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    hi
    \left(2x-1 \right)^{3} = \left(2x-1 \right)^{2} \times \left(2x-1 \right)
    study the sign of \frac{2x-1}{4x}
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  6. #6
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    Quote Originally Posted by mark1950 View Post
    Wow. thanks. How about if the numerator was (2x - 1)^3 as in always negative? What would the new solution set be? Thanks.

    If its  \frac{(2x-1)^3}{4x}<0

     \frac{(2x-1)^2(2x-1)}{4x}<0

    So you need to consider only \frac{2x-1}{4x}<0

    so the solution will be (0,1/2)
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