1. ## solve

2. Originally Posted by perash
$\displaystyle x=\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999 \sqrt{2000x}}}}}$

1. Square both sides
$\displaystyle x^{2}=x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\s qrt{2000x}}}}$

2. Subtract x from both sides
$\displaystyle x^{2}-x=1999\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{20 00x}}}}$

3. Divide by 1999
$\displaystyle \frac{x^{2}-x}{1999}=\sqrt{x+1999\sqrt{x+1999\sqrt{x+1999\sqrt {2000x}}}}$

4. Square both sides
$\displaystyle (\frac{x^{2}-x}{1999})^{2}=x+1999\sqrt{x+1999\sqrt{x+1999\sqrt{ 2000x}}}$

5. Simplify lefthand side
$\displaystyle \frac{x^{4}-2x^{3}+x^{2}}{1999^{2}}=x+1999\sqrt{x+1999\sqrt{x+ 1999\sqrt{2000x}}}$

6. Subtract x
$\displaystyle \frac{x^{4}-2x^{3}+x^{2}}{1999^{2}}-x=1999\sqrt{x+1999\sqrt{x+1999\sqrt{2000x}}}$

7. Common denominator and combine fractions
$\displaystyle \frac{x^{4}-2x^{3}+x^{2}-1999^{2}x}{1999^{2}}=1999\sqrt{x+1999\sqrt{x+1999\ sqrt{2000x}}}$

8. Divide by 1999
$\displaystyle \frac{x^{4}-2x^{3}+x^{2}-1999^{2}x}{1999^{3}}=\sqrt{x+1999\sqrt{x+1999\sqrt {2000x}}}$

9. Square both sides
$\displaystyle (\frac{x^{4}-2x^{3}+x^{2}-1999^{2}x}{1999^{3}})^{2}=x+1999\sqrt{x+1999\sqrt{ 2000x}}$

10. FOIL it out

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um, not going to finish, I just realized that this is annoying, and I am probably going to make a math error at some point, throwing the whole thing off. So I decided to look at the problem from a "is there an obvious answer" approach instead of a "how would I solve this" approach, and yes, the obvious answer is zero.

3. Suppose that $\displaystyle x=\sqrt{2000x}$ then $\displaystyle \sqrt{2000x} = \sqrt{x+1999x } = \sqrt{x+1999\sqrt{2000x}}$. Then $\displaystyle \sqrt{x+1999\sqrt{2000x}}=\sqrt{x+1999\sqrt{x+1999 x}}$ substitute again $\displaystyle x=\sqrt{x+1999\sqrt{x+1999\sqrt{2000x}}}$. Keep on going until you get what you have. So we need to solve $\displaystyle x=\sqrt{2000x}$ which has a trivial solution (as angel-white says) $\displaystyle x=0$ and a non-trivial solution $\displaystyle x=2000$.