
Re: Enrichment
Posted:
Aug 23, 2012 7:01 PM



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) + ((dbea)/a)x + (dcfa)/a
Now we can put this back into the quotient...
( (dx^2 + ex + f) + ((dbea)/a)x + (dcfa)/a ) / (dx^2 + ex + f)
= 1 + 1/a * ( (dbea)x + (dcfa)) / (dx^2 + ex + f)
We can drop the constants (1) and (1/a)...
((dbea)x + (dcfa)) / (dx^2 + ex + f)
And we can divide by (dbea)...
( x + (dcfa)/(dbea) ) / (dx^2 + ex + f)
At this point we have a term, (dcfa)/(bdea), 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 = (dcfa)/(dbea), so we don't have to keep writing this mess ...
(x + k) / (dx^2 + ex + f)
Let u = x + k ...
u / (d(uk)^2 + e(uk) + 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 = (dcfa)/(dbea)
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

