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Thread: question about radical problem

  1. #1
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    question about radical problem

    can someone tell me how 2x3+ 3squared + 2x3squared + 1 is 7+3x3squared? the only part i dont understand is where my book got the 3 from i thought it should be 7+2x3squared?
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by wonderd View Post
    can someone tell me how 2x3+ 3squared + 2x3squared + 1 is 7+3x3squared? the only part i dont understand is where my book got the 3 from i thought it should be 7+2x3squared?
    Hi wonderd,

    Try using parentheses for multiplication and ^ for taking a value to a power. It would help us tremendously.

    $\displaystyle 2(3)+3^2+2(3)^2+1 = 7+3(3)^2$

    $\displaystyle 6+9+18+1=7+27$

    $\displaystyle 34=34$
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  3. #3
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    this is the problem from the book, how do they get the 3?
    Attached Thumbnails Attached Thumbnails question about radical problem-untitled.bmp  
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  4. #4
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    Quote Originally Posted by wonderd View Post
    this is the problem from the book, how do they get the 3?
    Because $\displaystyle 2\sqrt3 + \sqrt3 = (2+1)\sqrt3$

    Just like $\displaystyle x + 2x = 3x$

    Your question adds $\displaystyle 2\sqrt3 to 1\sqrt3 = 3\sqrt3$
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  5. #5
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    y is 3sqaured 1?
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  6. #6
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    If you have two apples and someone gives you another apple how many do you have?

    Now imagine an apple is actually $\displaystyle \sqrt3$
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  7. #7
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    can someone explain this differently? bcuz one is not equal to 3 squared
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  8. #8
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    Thumbs down

    All they are doing is pretty much using the FOIL method, first let $\displaystyle x = \sqrt3 $

    $\displaystyle (\sqrt3 + 1)(2\sqrt3 + 1)$

    This can be re written as : $\displaystyle (x + 1)(2x + 1)$ by filling in x's for all square root 3's. now just FOIL that and you will get: $\displaystyle 2x^2 + x + 2x + 1$ combine like terms giving you: $\displaystyle 2x^2 + 3x + 1$ now fill in $\displaystyle \sqrt3$ for all the x's which gives you $\displaystyle 2(\sqrt3)^2 + 3\sqrt3 + 1$ Then once you simplify that, you get $\displaystyle 2(3) + 3\sqrt3 + 1 = 7 + 3\sqrt3$
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