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Math Help - Show Inequality X^2 > a

  1. #1
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    Show Inequality X^2 > a

    If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which -(sqrt{a}) < a (sqrt{a}).

    Okay...what exactly is the question asking in simple math terms?

    I don't follow the logic behind this sort of reasoning.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by symmetry View Post
    If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which -(sqrt{a}) < a (sqrt{a}).

    Okay...what exactly is the question asking in simple math terms?

    I don't follow the logic behind this sort of reasoning.
    First there is a typo in your post, the phrase:

    "consists of all numbers x for which -(sqrt{a}) < a (sqrt{a})"

    is probably mistyped as it is always true, there should be an x in there somewhere.

    If you sketch the graph of x^2-a (with a>0) you will see that this is >0 everywhere
    except between the roots of x^2-a, and when this is >0 we have x^2>a.

    So x^2>0 when x<-sqrt(a), or when x>sqrt(a).

    RonL
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  3. #3
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    Quote Originally Posted by symmetry View Post
    If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which -(sqrt{a}) < a (sqrt{a}).
    You have, for a>0.
    x^2 > a
    Express as,
    x^2 - a >0
    x^2 - (\sqrt{a})^2 > 0
    (x-\sqrt{a})(x+\sqrt{a})>0
    To be positive we require both factors to be positive or negative.

    1)Both positive. That means,
    x-\sqrt{a}>0
    x+\sqrt{a}>0
    Thus,
    x>\sqrt{a}
    x>-\sqrt{a}
    Another way of writing this is,
    x>\sqrt{a}
    Because if the top inequality is true then for certainly the bottom one, for it is contained in the top one.

    2)Both negativei. That means,
    x-\sqrt{a}<0
    x+\sqrt{a}<0
    Thus,
    x<\sqrt{a}
    x<-\sqrt{a}
    Another way of writing this is,
    x<-\sqrt{a}
    Because if the bottom inequality is true then for certainly the top one, for it is contained in the bottom one.

    Thus,
    x>\sqrt{a} \mbox{ or }x<-\sqrt{a}
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  4. #4
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    ok

    I want to thank both for your replies.

    Captainblack,

    You are right in making your statement.

    There is a typing error.

    I will now send the correct questions.

    Thanks.
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