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    i want to simplify: 2(root 7) + 8 / (root 7) + 2
    in the form p + q(root 7) where p and q are integers.

    can i simplify the above expression to: (root7) + 4

    is this correct?
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    (Original post by BABYGUY)
    i want to simplify: 2(root 7) + 8 / (root 7) + 2
    in the form p + q(root 7) where p and q are integers.

    can i simplify the above expression to: (root7) + 4

    is this correct?
    no
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    (Original post by BABYGUY)
    i want to simplify: 2(root 7) + 8 / (root 7) + 2
    in the form p + q(root 7) where p and q are integers.

    can i simplify the above expression to: (root7) + 4

    is this correct?
    I'm gonna get this done quickly because I want to go to bed

    2 root 7 + 8
    --------------
    root 7 + 2

    = (2root7+8)(root7-2)
    --------------------
    (root7+2)(root7-2)

    = (4root7 - 2)/3
    • Thread Starter
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    why????
    how can i make 2(root 7) = 8 / (root 7) + 2
    into form: p + q(root 7)

    please help.
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    yes but it says p + q(root 7)....your answer has 3 in a denominator, it doesnt want it to be a fraction.
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    (Original post by BABYGUY)
    yes but it says p + q(root 7)....your answer has 3 in a denominator, it doesnt want it to be a fraction.
    Then divide each bit by 3 to get 4/3 root 7 - 2/3.
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    (Original post by elpaw)
    -6 + 4/5 rt 7
    eh?
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    (Original post by theone)
    eh?
    sorry wrote the numbers down wrong
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    (Original post by theone)
    Then divide each bit by 3 to get 4/3 root 7 - 2/3.
    ...What he said.
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    Aha
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    -2/3 + 4/3 rt 7 is what i get (when i write the numbers down right! )
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    (Original post by BABYGUY)
    i want to simplify: 2(root 7) + 8 / (root 7) + 2
    in the form p + q(root 7) where p and q are integers.

    can i simplify the above expression to: (root7) + 4

    is this correct?
    thought you were 'too good for oxbridge' well, good luck
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    (Original post by DanMushMan)
    thought you were 'too good for oxbridge' well, good luck
    Hahaha! Sure told him!
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    (Original post by elpaw)
    -2/3 + 4/3 rt 7 is what i get (when i write the numbers down right! )
    Can you go through your working?

    p and q are supposed to be integers though???
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    (Original post by theone)
    Can you go through your working?
    you r correct
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    (Original post by Unregistered)
    p and q are supposed to be integers though???
    No - it's not possible. Stupid question by a stupid poster.
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    (Original post by theone)
    Can you go through your working?
    (8 + 2 rt 7)/(2 + rt 7) = ((8 + 2 rt 7)(2 - rt 7))/((2 + rt 7)(2 - rt 7))
    = (16 - 8 rt 7 + 4 rt 7 - 14)/(4 - 2 rt 7 + 2 rt 7 - 7)
    = (2 - 4 rt 7)/(-3)
    =-2/3 + 4/3 rt 7
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    (Original post by elpaw)
    (8 + 2 rt 7)/(2 + rt 7) = ((8 + 2 rt 7)(2 - rt 7))/((2 + rt 7)(2 - rt 7))
    = (16 - 8 rt 7 + 4 rt 7 - 14)/(4 - 2 rt 7 + 2 rt 7 - 7)
    = (2 - 4 rt 7)/(-3)
    =-2/3 + 4/3 rt 7
    <Puffs cheeks> Good stuff.
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    lol.........i love winding you up. i give these pointless maths threads where i already have the answer. and the funnire thing is, i lied about the integers thing, i said rational numbers lol..omg.you lot fell for not rationalising the denominator.hilarious. omg. funny. but i love hearing you lot squabble over these maths problems!!!! rember the last thread!! 10 pages long and i already knew the answer!!!
 
 
 

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