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Topic: Simple algebraic problem - but my young student rejects my method as too complex
Replies: 2   Last Post: Jan 6, 2014 11:31 PM

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@less@ndro

Posts: 215
Registered: 12/13/04
Re: Simple algebraic problem - but my young student rejects my method as too complex
Posted: Jan 6, 2014 10:24 PM
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Port563 <reader80@eternal-september.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 lower-powered 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 = (A-B)^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)((x-1)^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



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