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Thread: 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 $\displaystyle x\leq 2$ . Now, $\displaystyle \sqrt{x^2-6x+9}=\sqrt{(x-3)^2}=3-x$ , because $\displaystyle x\leq 2$ and it is supposed to choose the positive root . So, for all $\displaystyle x\leq 2$


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

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


    $\displaystyle \sqrt{x^2-6x+9}\;+\;\sqrt{2-x}\:+x-3=3-x+\;\sqrt{2-x}\;+x-3=\sqrt{2-x}$
    Sorry for the triviality of this question, but what's confusing me here is $\displaystyle \sqrt{(x-3)^2}=3-x$. Is it just true that $\displaystyle \sqrt{(x-3)^2}= \pm (3-x) $ ...but I still don't see what difference that makes because $\displaystyle \(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 $\displaystyle \sqrt{(x-3)^2}=3-x$. Is it just true that $\displaystyle \sqrt{(x-3)^2}= \pm (3-x) $ ...but I still don't see what difference that makes because $\displaystyle \(x-3)^2$ will be positive regardless of x.
    It is never true that $\displaystyle \sqrt{a^2}=\pm a$.

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

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

    If $\displaystyle x<3$ then $\displaystyle |x-3|=3-x$
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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 $\displaystyle \sqrt{a^2}=\pm a$.

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

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

    If $\displaystyle x<3$ then $\displaystyle |x-3|=3-x$
    (i) $\displaystyle \sqrt{9}=\pm 3$
    (ii) Consider the function $\displaystyle f(x)=\sqrt{x}$ . Then $\displaystyle f(9)=3$
    (iii) Consider the function $\displaystyle f(x)=-\sqrt{x}$ . Then $\displaystyle f(9)=-3$
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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 $\displaystyle \sqrt{(x-3)^2}=3-x$. Is it just true that $\displaystyle \sqrt{(x-3)^2}= \pm (3-x) $ ...but I still don't see what difference that makes because $\displaystyle \(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 $\displaystyle f(x)=\sqrt{g(x)}$ and $\displaystyle g(x)\geq 0$ we consider the corresponding positive root (or $\displaystyle 0$ ) and for $\displaystyle f(x)=-\sqrt{g(x)}$ we consider the corresponding negative root (or $\displaystyle 0$ ) . In our case $\displaystyle x\leq 2$ hence $\displaystyle 3-x> 0$ , for that reason $\displaystyle f(x)=\sqrt{(x-3)^2}=3-x$ (the corresponding positive root) .
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    Super Member ILikeSerena's Avatar
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    Re: Combining inequalities

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

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

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