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Re: Simple algebraic problem  but my young student rejects my method as too complex
Posted:
Jan 6, 2014 10:24 PM


Port563 <reader80@eternalseptember.org> wrote: > Find all real roots of: > > x^2 + 81x^2/(x+9)^2  40 = 0 > > The question is from one of the Greek wizard M.Skanavi's excellent problem > books (in Russian); at the level asked, no knowledge of the general solution > to cubics, let alone quartics, is assumed. > > Nothing cancels if you brute force it  one gets a clumsy polynomial > expression of degree 4, with all the lowerpowered terms too. > > Hence, substitution is the way. > > I solved it immediately by inspection of the 2nd term (x=0 is not a > solution, so I divided both its numerator and denominator by x) and making > the "obvious" substitution: > > y = 1 + 9/x > > and then, after simplification, a second substitution really cries out to be > made: > > z = y^2  y > > This yields a quadratic in z, with 2 real roots. > > Each yields a quadratic in y, but one of them has complex roots (yielding > complex x) and can be discarded. > > The other roots of y correspond to: > > x = 1 +/ SQRT(19) > > But he insists this is much harder than the problems in the set before and > after, and I am missing something simpler / prettier. > > Am I?
Perhaps previous problems had emphasized completing squares, so that your student expected something similar, like the following calculations:
x^2 + 81x^2/(x+9)^2  40
[ A = x , B = 9x/(x+9) , A^2 + B^2 40 = (AB)^2 +2AB  40 ]
= (x  9x/(x+9))^2 + 18x^2/(x+9)  40
[ simplifying first summand ]
= (x^2/(x+9))^2 + 18x^2/(x+9)  40
[ A = x^2/(x+9) , B = 9 , A^2 + 2AB 40 = (A+B)^2  81  40 ]
= (x^2/(x+9) + 9)^2  121
[ 3rd binomial ]
= (x^2/(x+9) + 20) (x^2/(x+9) 2)
= (x^2 + 20x + 180)(x^2  2x  18)/(x+9)^2
= ((x+10)^2 + 80)((x1)^2  19)/(x+9)^2
Now the solutions 1 +/ sqrt(19) can be read from the second factor. I have indicated the respective substituions above. I do not know if it looks prettier, but it could just be more familiar to your student.
 Marc



