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Re: Enrichment
Posted:
Aug 23, 2012 7:01 PM
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On Aug 20, 2012, at 5:29 PM, Dave L. Renfro <renfr1dl@cmich.edu> wrote:
> Jean Gaston Darboux, "Discussion de la fraction
> (ax^2 + bx + c)/(a'x^2 + b'x + c')
I couldn't leave this alone without taking a shot at the general quadratic quotient.
(ax^2 + bx + c) / (dx^2 + ex + f)
Following the same pattern I did with Mike's equation, I am going to put the top expression into a form containing the bottom expression as a term...
ax^2 + bx + c.
Multiply by d/a...
dx^2 + (db/a)x + dc/a
Add and subtract ex and f...
dx^2 + ex + f + (db/a)x - ex + dc/a - f
= (dx^2 + ex + f) + ((db-ea)/a)x + (dc-fa)/a
Now we can put this back into the quotient...
( (dx^2 + ex + f) + ((db-ea)/a)x + (dc-fa)/a ) / (dx^2 + ex + f)
= 1 + 1/a * ( (db-ea)x + (dc-fa)) / (dx^2 + ex + f)
We can drop the constants (1) and (1/a)...
((db-ea)x + (dc-fa)) / (dx^2 + ex + f)
And we can divide by (db-ea)...
( x + (dc-fa)/(db-ea) ) / (dx^2 + ex + f)
At this point we have a term, (dc-fa)/(bd-ea), that we didn't have with Mike's problem. We are going to use a couple substitutions to cajole this into the form we are seeking (a simple square)...
Let k = (dc-fa)/(db-ea), so we don't have to keep writing this mess ...
(x + k) / (dx^2 + ex + f)
Let u = x + k ...
u / (d(u-k)^2 + e(u-k) + f)
= u / ( du^2 - 2dku + dk^2 + eu - ek + f)
Take the reciprocal (like we did in Mike's problem) ...
( du^2 - 2dku + dk^2 + eu -ek + f ) / u
= du - 2dk + e + (dk^2 - ek + f ) / u
Drop the constant terms...
du + (dk^2 - ek + f) / u
divide by d
u + (k^2 - ek/d + f/d) / u
Let j = k^2 - ek/d + f/d ...
u + j/u
= (u^2 + j) / u
Essentially, we are now one step away from the same solution we had for Mike's equation...
Completing the square (note the plus or minus) ...
(u^2 + j) / u +/- 2sqrt(j)
= (u^2 +/- 2sqrt(j)*u + j) / u
= (u +/- sqrtj)^2 / u
= ((u +/- sqrt(j)) / sqrt(u)) ^ 2
A simple square and thus u = +/- sqrt(j)
Substituting x+k back in for u...
x = +/- sqrt(j) - k
where j = k^2 - ek/d + fd and k = (dc-fa)/(db-ea)
If you apply this to Mike's equation you find that k = 0 and j = 2 and the solution is x = +/- sqrt(2), I forgot the +/- in the solution to Mike's equation.
Bob Hansen
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