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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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    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

2 Answers 2

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If you don't mind trig parametrizations, another quite simple looking solution is:

$a = x - \tan(\phi)$

$d = \dfrac{2 g}{f \sin(2 \phi)}$

Maybe $\phi$ was the extra variable you mentioned, although the expression for $d$ doesn't involve $a$ explicitly.

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    Thanks for editing, J M - I tried "frac", but the result was too small. Must remember "dfrac" for future use ;-)2011-08-13
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This is probably not what you want, but here goes: $\displaylines{a=Q+17\cr d={((x-a)^2+1)g\over(x-a)f}\cr}$ Maybe if you tell us what's wrong with this answer, we can make some progress toward understanding your question.

EDIT: I wonder if what you really want is

new variable equals something containing $a$ but not $d$,

$d$ equals something containing new variable but not $a$

such as $Q=x-a$, $d=(Q^2+1)g/(Qf)$.