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Math Help - square roots

  1. #1
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    square roots

    It's a square root addition/subtraction but I don't know how to place the symbol for the square root i.e [ math ],

    so I'll try my best to describe it.

    square root of 3x^2 plus 6 square root of 108x^2 minus 4 square root of 108x^2

    Ok here is what I did;

    I subtracted + 6 square root of 108x^2 and -4 square root of 108x^2

    Then I squared off 108x^2 and got 12 square root of 3x^2 to which I added to the first square root. and I got with 12 square root of 3x^2

    Now the answer on the answer key is 13x square root of 3. I know I am wrong but I don't know where I am wrong.

    Does squaring off the 3x^2 give me 3 and the x on the outside?
    and what about the 13!?
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  2. #2
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    If you type [math] \sqrt {108x^2 } [/tex] then you will get \sqrt {108x^2 } .
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  3. #3
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    Hello, Morzilla!

    You did nothing wrong . . . just take another step.


    Simplify: . \sqrt{3x^2} + 6\sqrt{108x^2} - 4\sqrt{ 108x^2}

    we have: . \sqrt{3x^2} + \underbrace{6\sqrt{108x^2} - 4\sqrt{108x^2}}

    . . . . . . . . = \;\sqrt{3x^2} + 2\sqrt{108x^2}

    . . . . . . . . = \;\sqrt{3x^2} + 2\sqrt{36\!\cdot3x^2}

    . . . . . . . . = \;\sqrt{3x^2} + 2\!\cdot\!6\!\cdot\!\sqrt{3x^2}

    . . . . . . . . = \;1\!\cdot\!\sqrt{3x^2} + 12\!\cdot\!\sqrt{3x^2}

    . . . . . . . . = \;13\!\cdot\!\sqrt{3x^2} . . . .
    This is correct (so far)

    . . . . . . . . = \;13\!\cdot\!\sqrt{3}\!\cdot\!\sqrt{x^2}

    . . . . . . . . = \;13\!\cdot\sqrt{3}\!\cdot\!x

    . . . . . . . . = \;13x\sqrt{3}

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  4. #4
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    Smile A many, greateful thanks...

    Quote Originally Posted by Plato View Post
    If you type  \sqrt {108x^2 } then you will get \sqrt {108x^2 } .
    Quote Originally Posted by Soroban View Post
    Hello, Morzilla!

    You did nothing wrong . . . just take another step.



    we have: . \sqrt{3x^2} + \underbrace{6\sqrt{108x^2} - 4\sqrt{108x^2}}

    . . . . . . . . = \;\sqrt{3x^2} + 2\sqrt{108x^2}

    . . . . . . . . = \;\sqrt{3x^2} + 2\sqrt{36\!\cdot3x^2}

    . . . . . . . . = \;\sqrt{3x^2} + 2\!\cdot\!6\!\cdot\!\sqrt{3x^2}

    . . . . . . . . = \;1\!\cdot\!\sqrt{3x^2} + 12\!\cdot\!\sqrt{3x^2}

    . . . . . . . . = \;13\!\cdot\!\sqrt{3x^2} . . . .
    This is correct (so far)

    . . . . . . . . = \;13\!\cdot\!\sqrt{3}\!\cdot\!\sqrt{x^2}

    . . . . . . . . = \;13\!\cdot\sqrt{3}\!\cdot\!x

    . . . . . . . . = \;13x\sqrt{3}

    Many thanks plato!!!

    Ahhhhh ok, so there is a 1 in front of the square root!! that makes it 13!!! Thank you! that was part of my question as well!! After looking at it from the start again, I was not 100% sure if a 1 was supposed to be in front of  \sqrt {3x^2 }

    Also, is it me or the text edit options are gone?! the ones to change size, color, boldness and others!?

    Thanks
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