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Topic: Enrichment
Replies: 38   Last Post: Aug 24, 2012 1:33 PM

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Robert Hansen

Posts: 7,528
From: Florida
Registered: 6/22/09
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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