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Thread: [SOLVED] Solution Set

  1. #1
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    [SOLVED] Solution Set

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

    Hence we obtain a contradiction.

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

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

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

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

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

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

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

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

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