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Math Help - [SOLVED] Something bothering my understanding

  1. #1
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    [SOLVED] Something bothering my understanding



    The upper part is the question and lower part is my answer.
    Is a=-1, b=3, c=1 acceptable?
    Attached Thumbnails Attached Thumbnails [SOLVED] Something bothering my understanding-ques.jpg  
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  2. #2
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    Quote Originally Posted by stpmmaths View Post


    The upper part is the question and lower part is my answer.
    Is a=-1, b=3, c=1 acceptable?
    HI

    Yes , its acceptable as one of the answers . You will still need to test it and see whether it works fine with the matrix .

    \sqrt{1} = \pm 1 but in this case is -1 .

    However , if a=-1 , then its not acceptable .
    Last edited by mathaddict; October 28th 2009 at 12:36 AM.
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  3. #3
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    Quote Originally Posted by stpmmaths View Post


    The upper part is the question and lower part is my answer.
    Is a=-1, b=3, c=1 acceptable?
    Have you substituted a = 1, b = 3 and c = -1 into the matrix? Do you get a symmetric matrix?
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  4. #4
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    If I substituted a=1, b=3 and c=-1, I can't get a symmetric matrix. However by taking \sqrt{a} = -1, I can get a symmetric matrix.
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  5. #5
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    \sqrt{a} = -1
    By squaring both side,
    a = 1
    Is this true?

    But when we substitute back a = 1 into \sqrt{1} is not equal to -1.

    How?
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  6. #6
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    Quote Originally Posted by mathaddict View Post
    HI

    Yes , its acceptable as one of the answers . You will still need to test it and see whether it works fine with the matrix .

    \sqrt{1} = \pm 1 but in this case is -1 .

    However , if a=-1 , then its not acceptable .
    I don't understand this equation \sqrt{1} = \pm 1. I thought it should be \pm \sqrt{1} = \pm 1
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  7. #7
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    \sqrt{1}= \pm 1 is incorrect. \sqrt{1}= 1 since " \sqrt{a}" is defined as "the positive x such that x^2= a".

    It is true that the roots of the equation x^2= 1 are 1 and -1, but the reason we say "the roots of equation x^2= a are x= \pm\sqrt{a} is that " x= \sqrt{a}" does NOT give both roots.
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  8. #8
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    Quote Originally Posted by stpmmaths View Post
    I don't understand this equation \sqrt{1} = \pm 1. I thought it should be \pm \sqrt{1} = \pm 1
    gosh , i am wrong again . Sorry stpm math , Hallsofivy has corrected me .
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    \sqrt{1}= \pm 1 is incorrect. \sqrt{1}= 1 since " \sqrt{a}" is defined as "the positive x such that x^2= a".

    It is true that the roots of the equation x^2= 1 are 1 and -1, but the reason we say "the roots of equation x^2= a are x= \pm\sqrt{a} is that " x= \sqrt{a}" does NOT give both roots.
    So in this case, \sqrt{a} = -1 does not exist. Am I right?
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  10. #10
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    It is incorrect! I'm not sure I would say "it doesn't exist". That equation certainly does exist- it's right there on the screen!

    And both \sqrt{a} and -1 exist. They just can't be equal!
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  11. #11
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    So a=1, b=3 and c=-1 is not the answer. Right?
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  12. #12
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    Precisions about the question

    Hello,
    By finding the notation ( \sqrt{a}) in the text of your question, it is certain that your exercice concern the set of real numbers (not complex numbers) .
    The equality c = \sqrt{a} makes sence if and only if the variables a and c are defined as positive real numbers (includes the possible value of zero).
    So finding c=-1 makes the value of c out of defined range and thus there is no need to continue calculating the values of a and b.
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