# Finnish high school math problem.

1. Dec 12, 2008

### Lion_King

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. Dec 12, 2008

### mutton

With high exponents and common bases, the trick is usually to factor.

First, expand $$(2x)^{500}$$ and then move all terms to one side. Factor out $$x^2$$ 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.

3. Dec 12, 2008

### Lion_King

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...