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Finnish high school math problem.

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data

    x1000 = x(2x)500-2998x2


    Okay. This is difficult. It is easy to notice that the solutions are x=0 and x=2 but how do you prove that? I have tried to resolve it in different ways with no luck at all. Maybe I haven´t noticed something. Any ideas? This is Finnish final exam puzzle from 1989...
     
  2. jcsd
  3. Dec 12, 2008 #2
    With high exponents and common bases, the trick is usually to factor.

    First, expand [tex](2x)^{500}[/tex] and then move all terms to one side. Factor out [tex]x^2[/tex] and note that x = 0 is a solution. What remains is a quadratic, which can be factored to get x = 2 as the only other solution.
     
  4. Dec 12, 2008 #3
    How do you know that it is clever to expand (2x)500? For example, I tried to expand x2.

    Is this the correct way?
    x1000=x(2x)500-2998x2

    (2x)500x1000=(2x)500x(2x)500-(2x)5002998x2

    2500x1500-21000x1001+21498x502=0

    x2(2500x1498-21000x999+21498x500)=0
    Okay, clearly x = 0 is a solution. But what´s the difference? There is still that exponent monster. How can I get that x = 2? I am not following...
     
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