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Math Help - Adding/Sub Radicals

  1. #1
    Junior Member fluffy_penguin's Avatar
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    Arrow Adding/Sub Radicals

    Stuck...



    25\sqrt {3 } + 3\sqrt {3} - \sqrt {3}
    -----------------
    2 + 2 - 2

    Is the answer:

    28\sqrt {3}
    over 2
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  2. #2
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    Quote Originally Posted by fluffy_penguin View Post
    Stuck...



    25\sqrt {3 } + 3\sqrt {3} - \sqrt {3}
    -----------------
    2 + 2 - 2

    Is the answer:

    28\sqrt {3}
    over 2
    Well... Your math was slightly flawed.

    (5\sqrt {3 } + 3\sqrt {3} - \sqrt {3})/2

    Note the denominator is just the common divisor... not a sum/different/etc.

    Also note: \sqrt {75} equals \sqrt {5^2*3} equals 5\sqrt {3}
    This means you can factor out a 5 (the root of 25) and not 25.

    I should mention the answer...
    (5\sqrt {3 } + 3\sqrt {3} - \sqrt {3})/2
    equals
    ((5+3-1)\sqrt {3})/2
    (7\sqrt {3})/2
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  3. #3
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    \begin{array}{rcl} \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{{\sqrt 4 }} & = & \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{{\sqrt 4 }}\left( {\frac{{\sqrt 4 }}{{\sqrt 4 }}} \right) \\ <br />
  & = & \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt {12} }}{4} \\  \end{array}

    <br />
\begin{array}{rcl}<br />
  \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{{\sqrt 4 }}& = & \frac{{\sqrt {3 \cdot 25} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt {4 \cdot 3} }}{4} \\ <br />
  & = & \frac{{5\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{2\sqrt 3 }}{4} \\ <br />
 \end{array}
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  4. #4
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    Are you only taking positive roots? Because \frac{9\sqrt{3}}{2} is also
    a solution.
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  5. #5
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    Quote Originally Posted by SnipedYou View Post
    Are you only taking positive roots? Because \frac{9\sqrt{3}}{2} is also
    a solution.
    What in the world can that statement mean?
    What are you on about?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SnipedYou View Post
    Are you only taking positive roots? Because \frac{9\sqrt{3}}{2} is also
    a solution.
    Roots? There are no functions here. And there is only one answer for this problem.

    At a guess, perhaps you are confused about taking \pm when you take a square root? When you are looking at a numeric (that is to say, non-variable) expression involving a radical like a square root, you take the positive sign. This is a standard definition.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    At a guess, perhaps you are confused about taking \pm when you take a square root?
    Dan, what a nice guy you must be. That was such a polite way to put it.
    I wanted to say “mathematical amateurs ought to stay out of it.”
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    Dan, what a nice guy you must be. That was such a polite way to put it.
    I wanted to say “mathematical amateurs ought to stay out of it.”


    -Dan
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