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Math Help - Combining inequalities

  1. #1
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    Combining inequalities

    1st post here.

    I came across this problem in a GMAT reviewer and I naturally got B, but the correct answer is listed as A.

    There was a brief explanation about why the answer is A: (2-x)^(1/2), but the brief explanation confused me. I realize that x>=2, but I don't see how I can use that to get the correct answer.
    Attached Thumbnails Attached Thumbnails Combining inequalities-screen-shot-2011-12-18-5.26.21-pm.png  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Combining inequalities

    Quote Originally Posted by WSox24 View Post
    I realize that x>=2
    It should be x\leq 2 . Now, \sqrt{x^2-6x+9}=\sqrt{(x-3)^2}=3-x , because x\leq 2 and it is supposed to choose the positive root . So, for all x\leq 2


    \sqrt{x^2-6x+9}\;+\;\sqrt{2-x}\:+x-3=3-x+\;\sqrt{2-x}\;+x-3=\sqrt{2-x}
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    Re: Combining inequalities

    Quote Originally Posted by FernandoRevilla View Post
    It should be x\leq 2 . Now, \sqrt{x^2-6x+9}=\sqrt{(x-3)^2}=3-x , because x\leq 2 and it is supposed to choose the positive root . So, for all x\leq 2


    \sqrt{x^2-6x+9}\;+\;\sqrt{2-x}\:+x-3=3-x+\;\sqrt{2-x}\;+x-3=\sqrt{2-x}
    Sorry for the triviality of this question, but what's confusing me here is \sqrt{(x-3)^2}=3-x. Is it just true that \sqrt{(x-3)^2}= \pm (3-x) ...but I still don't see what difference that makes because \(x-3)^2 will be positive regardless of x.

    I guess I need to review some basic properties of abs value, roots, and inequalities
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    Re: Combining inequalities

    Quote Originally Posted by WSox24 View Post
    Sorry for the triviality of this question, but what's confusing me here is \sqrt{(x-3)^2}=3-x. Is it just true that \sqrt{(x-3)^2}= \pm (3-x) ...but I still don't see what difference that makes because \(x-3)^2 will be positive regardless of x.
    It is never true that \sqrt{a^2}=\pm a.

    This is true \sqrt{a^2}=|a|\ge 0.

    Thus \sqrt{(x-3)^2}=|3-x|.

    If x<3 then |x-3|=3-x
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Combining inequalities

    Quote Originally Posted by Plato View Post
    It is never true that \sqrt{a^2}=\pm a.

    This is true \sqrt{a^2}=|a|\ge 0.

    Thus \sqrt{(x-3)^2}=|3-x|.

    If x<3 then |x-3|=3-x
    (i) \sqrt{9}=\pm 3
    (ii) Consider the function f(x)=\sqrt{x} . Then f(9)=3
    (iii) Consider the function f(x)=-\sqrt{x} . Then f(9)=-3
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Combining inequalities

    Quote Originally Posted by WSox24 View Post
    Sorry for the triviality of this question, but what's confusing me here is \sqrt{(x-3)^2}=3-x. Is it just true that \sqrt{(x-3)^2}= \pm (3-x) ...but I still don't see what difference that makes because \(x-3)^2 will be positive regardless of x.

    I guess I need to review some basic properties of abs value, roots, and inequalities
    There is an universal mathematical convention. For the function f(x)=\sqrt{g(x)} and g(x)\geq 0 we consider the corresponding positive root (or 0 ) and for f(x)=-\sqrt{g(x)} we consider the corresponding negative root (or 0 ) . In our case x\leq 2 hence 3-x> 0 , for that reason f(x)=\sqrt{(x-3)^2}=3-x (the corresponding positive root) .
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  7. #7
    Super Member ILikeSerena's Avatar
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    Re: Combining inequalities

    Quote Originally Posted by FernandoRevilla View Post
    (i) \sqrt{9}=\pm 3
    From wikipedia:
    "Every non-negative real number x has a unique non-negative square root, called the principal square root, which is denoted by \sqrt, where \sqrt is called radical sign."

    I'm not aware of any convention where it is possible that \sqrt{9}=-3.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Combining inequalities

    Quote Originally Posted by ILikeSerena View Post
    "Every non-negative real number x has a unique non-negative square root,
    Right, and a unique non-positive square root.

    I'm not aware of any convention where it is possible that \sqrt{9}=-3.
    Neither do I.
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  9. #9
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    Re: Combining inequalities

    Quote Originally Posted by FernandoRevilla View Post
    (i) \sqrt{9}=\pm 3
    I suppose that depends upon one's mathematics education community.
    In North America we stress that \sqrt{a}\ge 0, if defined.
    That includes the GMAT test.
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