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    Can someone please help me rearrange this equation? Also what will I be writing in the 4th column? Will it be square root f? Attachment 477271


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    (Original post by Clarehd1310)
    Can someone please help me rearrange this equation? Also what will I be writing in the 4th column? Will it be square root f? Attachment 477271


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    Do (b) (i) first and list f in the fourth column, then move on to (b) (ii) and re-arrange the equation to

    \sqrt{f} = \sqrt{B}Z - \sqrt{B} where y = \sqrt{f}, m=\sqrt{B} and c = -\sqrt{B} as required.
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    Since it's been square rooted just square it and solve for f.

    So, we start with (by the way x = 'multiply', its not a variable):
    Root(f)=Root(B) x (Z-1) <<<< square everything
    f = B x (Z-1)^2 <<<<< this is the same as saying
    f = B x (Z-1)x(Z-1) <<<<< now expand the brackets
    f = B x (Z^2 - 2Z + 1)
    f = BZ^2 - 2Z + 1

    Done, since the question wants Z as the x variable, this looks like a normal quadratic to me. Plug in the B value you found in the last equation if you like.
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    (Original post by Omar J to the T)
    Since it's been square rooted just square it and solve for f.

    So, we start with (by the way x = 'multiply', its not a variable):
    Root(f)=Root(B) x (Z-1) <<<< square everything
    f = B x (Z-1)^2 <<<<< this is the same as saying
    f = B x (Z-1)x(Z-1) <<<<< now expand the brackets
    f = B x (Z^2 - 2Z + 1)
    f = BZ^2 - 2Z + 1

    Done, since the question wants Z as the x variable, this looks like a normal quadratic to me. Plug in the B value you found in the last equation if you like.
    This isn't correct.
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    (Original post by Zacken)
    This isn't correct.
    So whats the answer?
 
 
 
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