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Math Help - Radicals with different Radicands

  1. #1
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    Radicals with different Radicands

    Hello all, Im posting here for the first time and need a little help with some work.

    Im trying to work on this problem

    <br />
5\sqrt{72} - \sqrt{32} + 6\sqrt{98}<br />
    but im absolutely lost....

    Thanks for the help in advance,
    Aaron
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  2. #2
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    I can help you buy how do you type it all nice like that? with the square root and everything i mean.
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  3. #3
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    type like the

    math]
    5\sqrt{72} - \sqrt{32} + 6\sqrt{98}
    /math]

    then just put the brackets on to complete the code to make it look snazzy
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  4. #4
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    [tex]
    \sqrt{25*72} - \sqrt{32} + \sqrt{36*98}
    /[tex]
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  5. #5
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    try putting the forward slash in the bottom bracket to make it look like [/tex]

    <br />
\sqrt{25*72} - \sqrt{32} + \sqrt{36*98}<br />
    Im not sure if the equation above is correct, because according ot my math book it says "These radicals cannot be be subtracted in their present form because they contain different radicands. When that occurs, determine whether one or more of the radicals can be simplified so that they have the same radicands. then it gives an example of
    <br />
5\sqrt{3} - \sqrt{12} = 5\sqrt{3} - \sqrt{4*3}
     = 5\sqrt{3} - \sqrt{4} * \sqrt{3}
    = 5\sqrt{3} - 2\sqrt{3}
     = (5 - 2)\sqrt{3} = 3\sqrt{3}
    But the book doesnt give an example of how to do both adding and subtracting radicals with different radicands
    Last edited by Alundblade; August 30th 2009 at 09:07 PM.
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  6. #6
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    heres the answer
    [/tex]
    68\sqrt{2}
    [/tex]
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  7. #7
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    for god sakes 68 times the square root of 2 lol

    If you want me to show you how i got it ask me.
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  8. #8
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    well thank you for the answer but I need to know how it was solved so i can continue working on the rest of the quiz
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  9. #9
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    Quote Originally Posted by usmelikchees View Post
    for god sakes 68 times the square root of 2 lol

    If you want me to show you how i got it ask me.
    How did you get it?
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  10. #10
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    Ill explain in words the best I can.

    EX) 5*sqrt2 is the same thing at sqrt(2*25)

    This is done by squaring the number that the square root is being multiplied by in this case the 5.

    So sqrt of (2*25) or 50 is 5*sqrt2.
    __________________________________________
    What I did to solve your problem was make all the parts of the problem to be multiplied by the same square root (sqrt2)

    Need further explaining?
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  11. #11
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    Quote Originally Posted by Alundblade View Post
    try putting the forward slash in the bottom bracket to make it look like [/tex]

    <br />
\sqrt{25*72} - \sqrt{32} + \sqrt{36*98}<br />
    Im not sure if the equation above is correct, because according ot my math book it says "These radicals cannot be be subtracted in their present form because they contain different radicands. When that occurs, determine whether one or more of the radicals can be simplified so that they have the same radicands. then it gives an example of
    <br />
5\sqrt{3} - \sqrt{12} = 5\sqrt{3} - \sqrt{4*3}
     = 5\sqrt{3} - \sqrt{4} * \sqrt{3}
    = 5\sqrt{3} - 2\sqrt{3}
     = (5 - 2)\sqrt{3} = 3\sqrt{3}
    But the book doesnt give an example of how to do both adding and subtracting radicals with different radicands
    The equation couldnt be added or subttracted in its present form so i changed it so they all had the same square root.
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  12. #12
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    alright thanks!
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