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  1. #1
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    algebra

    if  \frac{a}{b}=\frac{1+p}{1-p} , find the value of \frac{a^2+b^2}{a^2-b^2} in terms of p.
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  2. #2
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    Hello,
    Quote Originally Posted by requal View Post
    if  \frac{a}{b}=\frac{1+p}{1-p} , find the value of \frac{a^2+b^2}{a^2-b^2} in terms of p.
    \frac{a^2+b^2}{a^2-b^2}=\frac{b^2 \left(\frac{a^2}{b^2}+1\right)}{b^2 \left(\frac{a^2}{b^2}-1\right)}=\frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}

    Better now ?


    Otherwise, you could write a=b \cdot \frac{1+p}{1-p} and just substitute. There would have been a common factor : b^2
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  3. #3
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    Hello, requal!

    Another approach . . .


    \text{If }\,\frac{a}{b}=\frac{1+p}{1-p},\:\text{ find the value of: }\:\frac{a^2+b^2}{a^2-b^2}\,\text{ in terms of }p.
    We have: . \frac{a}{b} \:=\:\frac{1+p}{1-p} \quad\Rightarrow\quad \frac{a^2}{b^2} \:=\:\frac{1+2p + p^2}{1-2p+p^2} .[1]


    Add 1 to both sides of [1]: . \frac{a^2}{b^2}+1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} + 1

    . . and we have: . \frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2} \quad\Rightarrow\quad a^2+b^2\:=\:\frac{2b^2(1+p^2)}{(1-p)^2} .[2]


    Subtract 1 from both sides of [1]: . \frac{a^2}{b^2}-1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} - 1

    . . and we have: . \frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2} \quad\Rightarrow\quad a^2-b^2 \:=\:\frac{4b^2p}{(1-p)^2} .[3]


    Divide [2] by [3]: . \frac{a^2+b^2}{a^2-b^2} \;=\;\frac{\dfrac{2b^2(1+p^2)}{(1-p)^2}} {\dfrac{4b^2p}{(1-p)^2}} \;=\;\frac{1+p^2}{2p}\quad\hdots\:\text{ta-}DAA!

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  4. #4
    Moo
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    It's pretty equivalent

    Maybe yours can still be simplified.

    Quote Originally Posted by Soroban View Post
    . . and we have: . \frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2} \quad\Rightarrow\quad a^2+b^2\:=\:\frac{2b^2(1+p^2)}{(1-p)^2} .[2]
    Just keep \frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2}

    . . and we have: . \frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2} \quad\Rightarrow\quad a^2-b^2 \:=\:\frac{4b^2p}{(1-p)^2} .[3]
    Keep \frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2}

    Dividing the two equations, we directly find \frac{a^2+b^2}{a^2-b^2}


    The problem is that maybe one can't see why we have to subtract/add 1 to aČ/bČ
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,

    \frac{a^2+b^2}{a^2-b^2}=\frac{b^2 \left(\frac{a^2}{b^2}+1\right)}{b^2 \left(\frac{a^2}{b^2}-1\right)}=\frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}

    Better now ?
    Sorry to bump this thread but I don't follow? I was thinking you were trying to make it into 1+p/1-p but its thats in the form of 1+p/p+1?
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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, requal!

    Another approach . . .


    We have: . \frac{a}{b} \:=\:\frac{1+p}{1-p} \quad\Rightarrow\quad \frac{a^2}{b^2} \:=\:\frac{1+2p + p^2}{1-2p+p^2} .[1]


    Add 1 to both sides of [1]: . \frac{a^2}{b^2}+1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} + 1

    . . and we have: . \frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2} \quad\Rightarrow\quad a^2+b^2\:=\:\frac{2b^2(1+p^2)}{(1-p)^2} .[2]


    Subtract 1 from both sides of [1]: . \frac{a^2}{b^2}-1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} - 1

    . . and we have: . \frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2} \quad\Rightarrow\quad a^2-b^2 \:=\:\frac{4b^2p}{(1-p)^2} .[3]


    Divide [2] by [3]: . \frac{a^2+b^2}{a^2-b^2} \;=\;\frac{\dfrac{2b^2(1+p^2)}{(1-p)^2}} {\dfrac{4b^2p}{(1-p)^2}} \;=\;\frac{1+p^2}{2p}\quad\hdots\:\text{ta-}DAA!

    Is there some special reason that you would add/subtract one from both equations? Or is it just to make it easier to do
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  7. #7
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    Quote Originally Posted by Moo View Post
    Hello,

    \frac{a^2+b^2}{a^2-b^2}=\frac{b^2 \left(\frac{a^2}{b^2}+1\right)}{b^2 \left(\frac{a^2}{b^2}-1\right)}=\frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}

    Better now ?
    Quote Originally Posted by requal View Post
    Sorry to bump this thread but I don't follow? I was thinking you were trying to make it into 1+p/1-p but its thats in the form of 1+p/p+1?
    It is not bumping when you ask a follow-up question. Follow-up questions are always welcome and encouraged.

    Substitute \frac{a}{b} = \frac{1 + p}{1 - p} into \frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}.
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