喵喵喵 XVI

Problem 1.

For small values of \(|x|\), \(\sqrt{x+2026}\approx45\). Also, this is where \(x^2-2026\) is small, so we want \(x^2-2026\approx45\) as a good approximation to the roots. Solving this gives \(x'_1\gtrsim45.5,x_2'\lesssim-45.5\). (Notice that \(45.5^2=2025+45+\frac14=2070\frac14<2026+45\).)

We deal with \(x'_1\) first. We know that the quartic cannot have rational roots because every factor of \(2025\times2026\) (which is the constant) near \(\pm45\) doesn't give a solution. Since \(x_1\approx45+\frac12\), we might try \((x-\frac12)^2\approx2025\) which is \(x^2-x+\frac14\approx2025\). Trialing the constant term (could only be \(2025\) or \(2026\)) and we get a root \(x_1=\frac{1+\sqrt{8105}}2\) from the polynomial \(x^2-x-2026\). From there we can also extract \(x_2=\frac{1-\sqrt{8105}}2\).

We also know that the quartic has a \(x^3\) coefficient of \(0\) and a constant term of \(2025\times2026\). That means the other quadratic that gives the root has to be \(x^2+x-2025\) by Vieta's theorem. Therefore the third and fourth roots are \(x_3=\frac{-1+\sqrt{8101}}2,x_4=\frac{-1-\sqrt{8101}}2\).

\(x_2\) and \(x_3\) are discarded because \(x^2-2026>0\) which implies \(|x|>45\). Therefore the answer to this problem is \(\boxed{\frac{1+\sqrt{8105}}2\text{ and }\frac{-1-\sqrt{8101}}2}\)