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Re: Enrichment
Posted:
Aug 23, 2012 7:05 PM
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The second to the last line "where j = k^2 - ek/d + fd and k = (dc-fa)/(db-ea)" should be...
where j = k^2 - ek/d + f/d and k = (dc-fa)/(db-ea)
The j expression was mistyped.
On Aug 23, 2012, at 7:01 PM, Robert Hansen <bob@rsccore.com> wrote:
>
> 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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