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Math Help - Inequality problem.

  1. #1
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    Inequality problem.

    Solve the below inequality :-

    <br />
{(x^2 - 2x - 3)}^{1/2} < 1<br />

    Any help would be appreciated.

    Thanks
    Last edited by a69356; June 3rd 2009 at 07:17 AM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by a69356 View Post
    Solve the below inequality :-

    (x^2 - 2x - 3)^(1/2) < 1

    Any help would be appreciated.

    Thanks
    The square root of any number larger than 1 is itself larger than 1.
    That means that to have square root less than 1, the number itself must be less than 1.

    So we have to solve x^2- 2x- 3< 1 which is the same as x^2- 2x< 4.

    Adding 1 to both sides, to "complete the square", x^2- 2x+ 1< 5 which is the same (x-1)^2< 5. Now, (x-1)^2= 5 when x- 1= \pm\sqrt{5} or x= 1\pm\sqrt{5}. Since x= 0, which is between those values, satisfies the inequality, all numbers between them and only numbers between them satisfy the inequality. (For continuous functions, inequalities can only reverse at points where they are [b]equalties.)
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by a69356 View Post
    Solve the below inequality :-

    <br />
{(x^2 - 2x - 3)}^{1/2} < 1<br />

    Any help would be appreciated.

    Thanks
    HELLO :
    they have :

    an important condition is :

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