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Math Help - why always?

  1. #1
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    why always?

    When I do problems involving real roots, such as x = 1, why is it always x = 1 or x = - 1?

    For example:

    Determine the real roots of each equation:
    (5x^2+20)(3x^2-48)=0

    the real root is x = 4 or x = - 4. Why is it always +/- or how can I identify this issue?
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  2. #2
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    Quote Originally Posted by Barthayn View Post
    When I do problems involving real roots, such as x = 1, why is it always x = 1 or x = - 1?

    For example:
    Determine the real roots of each equation:
    (5x^2+20)(3x^2-48)=0
    the real root is x = 4 or x = - 4. Why is it always +/- or how can I identify this issue?
    But of course that is not always true for all roots. It is true for square roots.
    There are two real square roots of 16, (4)^2=16~\&~(-4)^2=16
    There are no real square roots of -16

    There is only one real cube root of 8, 2^3=8.
    There are no others and we did not use \pm.
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  3. #3
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    So basically if you can just square root it it will be +/- always? If it is cubic you can only have one root?
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  4. #4
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    Quote Originally Posted by Barthayn View Post
    So basically if you can just square root it it will be +/- always? If it is cubic you can only have one root?
    It is very dangerous to answer a question that contains 'always'.
    But yes that is generally true for real numbers.
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  5. #5
    Senior Member pacman's Avatar
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    it is because you generated a factor of the form a^2 - b^2 = (a + b)(a - b), the difference of two squares. And why not? They are just exercises, training wheels for a rigorous future math scrimages, if you mind. And multiplicity of roots often comes out handy with some other textbooks . . . .
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