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    How would you simplify
    4 times (root 11)/5+(root 11).
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    (Original post by Maths1210)
    How would you simplify
    4 times (root 11)/5+(root 11).
    You would try to simplify by rationalising the denominator. So you would multiply top and bottom by the conjugate of 5+(root 11).
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    (Original post by Anonymouspsych)
    You would try to simplify by rationalising the denominator. So you would multiply top and bottom by the conjugate of 5+(root 11).
    Okay. Then how would I be able to work out 4 (root 11) x 5+(root 11)?
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    (Original post by Maths1210)
    Okay. Then how would I be able to work out 4 (root 11) x 5+(root 11)?
    You want to multiply by the conjugate of 5+(root 11) which would be 5-(root 11). This means the top would turn into 4 (root 11) x 5-(root 11) and the bottom would turn into [5-(root 11)] x [5+(root 11)] .


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    (Original post by Anonymouspsych)
    You want to multiply by the conjugate of 5+(root 11) which would be 5-(root 11). This means the top would turn into 4 (root 11) x 5-(root 11) and the bottom would turn into [5-(root 11)] x [5+(root 11)] .

    Okay I still don't know where to go from there though.
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    (Original post by Maths1210)
    Okay I still don't know where to go from there though.
    would the 5 minus root 11 cancel out
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    (Original post by Maths1210)
    would the 5 minus root 11 cancel out
    do you not know how to expand two brackets with two terms?

    For example the bottom part of the fraction would turn into

    (5+root11)(5-root11) = 25 -5root11 + 5root11 - 11 = 14

    notice how the root11s cancel out- this is knows as the difference of two squares.

    Do the same technique with the top of the fraction and then simplify
 
 
 

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