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Math Help - Absolute value inequalities - I'm stuck...

  1. #1
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    Absolute value inequalities - I'm stuck...

    Hi - I hoped someone maybe able to help me understand this question:
    absolutevalue\(\frac{(x+1)}{(x-3)(x-5)}) < \frac{1}{2}

    (apologies but I can't find the syntax for absolute value sign)

    I know that I would need to square both sides in order to remove the modulus. But when rearranging I dont end up with a product so unsure how to proceed.

    Thanks, F
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  2. #2
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    Re: Absolute value inequalities - I'm stuck...

    Quote Originally Posted by FelixHelix View Post
    Hi - I hoped someone maybe able to help me understand this question:
    \left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}

    I know that I would need to square both sides in order to remove the modulus. But when rearranging I dont end up with a product so unsure how to proceed.
    LaTeX Note: [tex]\left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}[/tex] gives
    \left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}

    If you do square.
    \frac{(x+1)^2}{(x-3)^2(x-5)^2} < \frac{1}{4}.

    4(x+1)^2<(x-3)^2(x-5)^2.

    Be careful: x\ne 3~\&~x\ne 5
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  3. #3
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    Re: Absolute value inequalities - I'm stuck...

    Quote Originally Posted by Plato View Post
    LaTeX Note: [tex]\left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}[/tex] gives
    \left|\frac{(x+1)}{(x-3)(x-5)}\right| < \frac{1}{2}

    If you do square.
    \frac{(x+1)^2}{(x-3)^2(x-5)^2} < \frac{1}{4}.

    4(x+1)^2<(x-3)^2(x-5)^2.

    Be careful: x\ne 3~\&~x\ne 5
    Thanks for the latex tip.

    So with what you say in mind. If I draw a number line x cannot be equal to 5 or 3 but can be -1. And it will be always be positive between critical values because of the squaring of each parenthesis. Therefore the answers will be:

    x\le-1, -1< x <3, 3< x <5, x > 5

    Is this correct?
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  4. #4
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    Re: Absolute value inequalities - I'm stuck...

    Quote Originally Posted by FelixHelix View Post
    x\le-1, -1< x <3, 3< x <5, x > 5
    Is this correct?
    That nowhere close to correct!
    For one, x=2 is not a solution.
    Is x=6~?
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