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Math Help - inequality

  1. #1
    Super Member 11rdc11's Avatar
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    inequality

    How do I solve this?

    .01 > \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+1}}

    I just can't seem to figure it out, thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    How do I solve this?

    .01 > \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+1}}

    I just can't seem to figure it out, thanks
    you want \frac 1{\sqrt {x}} - \frac 1{\sqrt{x + 1}} - 0.01 < 0

    \Rightarrow \frac {\sqrt{x + 1} - 1 - 0.01 \sqrt{x} \sqrt{x + 1}}{\sqrt{x(x + 1)}} < 0

    now, for the fraction to be less than zero, we need the numerator and the denominator to be different signs. so, WLOG, just solve for

    \sqrt{x + 1} - 1 - 0.01 \sqrt{x} \sqrt{x + 1} < 0

    and

    \sqrt{x(x + 1)} > 0

    when you find the required x-values, plot them on a number line, and test values in all the regions to see if the satisfy the original inequality, this will give you your intervals of solutions
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  3. #3
    Super Member 11rdc11's Avatar
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    Thanks a million

    {\sqrt{x(x + 1)}} = \sqrt{x}\sqrt{x+1}

    Guess I got to work on my algebra

    Quick question though shouldn't the numerator be

    {\sqrt{x + 1} - \sqrt{x} - 0.01 \sqrt{x} \sqrt{x + 1}}
    Last edited by 11rdc11; September 3rd 2008 at 08:46 PM.
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  4. #4
    Super Member 11rdc11's Avatar
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    I'm still having problems solving the numerator. Can someone show me how to do it?

    I use ln to simply it but still a bit confused.
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  5. #5
    Moo
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    Hello,
    Quote Originally Posted by 11rdc11 View Post
    Quick question though shouldn't the numerator be

    {\sqrt{x + 1} - \sqrt{x} - 0.01 \sqrt{x} \sqrt{x + 1}}
    Yes it should.

    Ok, here are the steps I've used, for the brute force method

    \sqrt{x+1}<\sqrt{x}+0.01 \sqrt{x} \sqrt{x+1}

    Square the inequation (it is possible because both sides are positive). There is an x that will cancel out.
    After that, group all the terms that don't contain any square root in one side, and the stuff with the square root the other side.
    Let X=x(x+1) and then, square again.

    You'll have a quadratic inequality. Remember that it's the opposite sign of a within the interval formed by the roots.

    Then, you'll have to solve twice a quadratic.... well, you'll see ^^'
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