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Math Help - Adding Square Roots

  1. #1
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    Adding Square Roots

    i need to add 5 square root (7) and 3 square root (28). i can't figure out how to do it. please help me!!!!!
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  2. #2
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    Quote Originally Posted by EmmGee View Post
    i need to add 5 square root (7) and 3 square root (28). i can't figure out how to do it. please help me!!!!!
    First of all, the square root of 28= sq.root(7*4).....so it is also equal to 2 square root 7....so... 3 root 28 = 6 root 7.

    so, 5 square root 7 + 6 square root 7 = 11 square root 7.

    I dont know if my math is correct, but you just have to simplify your square roots.

    Sorry for the sloppy math....
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by blair_alane View Post
    First of all, the square root of 28= sq.root(7*4).....so it is also equal to 2 square root 7....so... 3 root 28 = 6 root 7.

    so, 5 square root 7 + 6 square root 7 = 11 square root 7.

    I dont know if my math is correct, but you just have to simplify your square roots.

    Sorry for the sloppy math....
    It is correct
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  4. #4
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    In general, you can break square roots apart - for example, \sqrt{44} = \sqrt{4}\times\sqrt{11}. Of course, the square root of 4 is 2, so you get 2\sqrt{11}.

    When you look to break apart larger numbers under a square root sign, you want to try to break them up into two factors where one of the factors is a perfect square (4, 9, 16, ...).
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  5. #5
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    yes! the first one I got right! here...I'll write it again...

    28=\sqrt(7*4).....so it is also equal to 2\sqrt7....so...
    3\sqrt28=3*2\sqrt7=6\sqrt7

    so, 5\sqrt7+6\sqrt7=11\sqrt7

    is that somewhat less confusing? (i found out how to do latex! teehee) I probably just made it worse! sorry! I dont know how to lay it out....

    hope that helps!
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  6. #6
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    Quote Originally Posted by Mathnasium View Post
    In general, you can break square roots apart - for example, \sqrt{44} = \sqrt{4}\times\sqrt{11}. Of course, the square root of 4 is 2, so you get 2\sqrt{11}.

    When you look to break apart larger numbers under a square root sign, you want to try to break them up into two factors where one of the factors is a perfect square (4, 9, 16, ...).
    you are totally right!
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by blair_alane View Post
    yes! the first one I got right! here...I'll write it again...

    28=\sqrt(7*4).....so it is also equal to 2\sqrt7....so...
    3\sqrt28=3*2\sqrt7=6\sqrt7

    so, 5\sqrt7+6\sqrt7=11\sqrt7

    is that somewhat less confusing? (i found out how to do latex! teehee) I probably just made it worse! sorry! I dont know how to lay it out....

    hope that helps!
    type [tex]\sqrt{Anything}[/tex] to get \sqrt{Anything}.

    so it would look nicer if you had \sqrt{7 * 4} or better yet \sqrt{7 \cdot 4} ..........use \cdot to get the dot. so the last one we got by typing [tex]\sqrt{7 \cdot 4}[/tex]
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    type [tex]\sqrt{Anything}[/tex] to get \sqrt{Anything}.

    so it would look nicer if you had \sqrt{7 * 4} or better yet \sqrt{7 \cdot 4} ..........use \cdot to get the dot. so the last one we got by typing [tex]\sqrt{7 \cdot 4}[/tex]
    teeheehee! Thank you, Jhevy! I shall remember this!

    Anyway, I hoped that helped EmmGee!
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