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Assuming $f\neq 0$ I'm trying to rewrite $$0 = 2 \cdot g \cdot ((x-a)^2 + 1) - 2 \cdot d \cdot f \cdot (x - a)$$

into a system of equations of the form

$a = $(something not containing $d$)

$d = $(something with $a$ in it)

I'm pretty sure you need to introduce another variable but I'm just not seeing the "trick".

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    You won't get a linear system, and I don't think you can decouple the variables, but here goes: divide by 2, put $df(x-a)$ on the left side, divide out the extra factor to get $d=\cdots$; write $y=x-a$ and solve the equation as a quadratic equation to get $y$ in terms of $g, f, d$, then write $a=x-y$.2011-07-13
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    Thank you for the comments, but I really would like the variables decoupled. Eventually I need to generalize for more terms and the quadratic trick won't work for this.2011-07-13
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    Are you saying you want the upper equation not to contain $d$ and/or the lower equation not to contain $a$? That won't be possible. You only have one scalar equation, you can't turn it into two independent ones. Are you thinking of something like $xy=0\Rightarrow x=0\lor y=0$? There's an "or" in there, whereas between your two equations there's an implied "and".2011-07-13
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    If that isn't possible, maybe relax the condition so $d = $(something with $a$ in it) but $a = $(something not containing $d$)2011-07-13
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    Since you can solve the quadratic for $a$ and get an expression for $a$ that depends on $d$, it's impossible to also get an expression for $a$ that doesn't depend on $d$.2011-07-13

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