
Re: Enrichment
Posted:
Aug 23, 2012 7:05 PM



The second to the last line "where j = k^2  ek/d + fd and k = (dcfa)/(dbea)" should be...
where j = k^2  ek/d + f/d and k = (dcfa)/(dbea)
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) + ((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 > > >

