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Home » $ x + u = 2 $, $ y + v = 4 $, - Dyverse

$ x + u = 2 $, $ y + v = 4 $, - Dyverse

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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x^4 + 1 = x^4 + \frac{1}{x^4} \cdot x^4 = x^4 + 1 \Rightarrow \text{No...
Instead, use:
\frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1} = \frac{(x^2 + 1)^2...
Let’s compute $ x^4 + 1 $, $ x^4 - 1 $:
We know $ x^2 + \frac{1}{x^2} = 23 \Rightarrow x^4 + \frac{1}{x^4} = 5...
x^4 + 1 = x^4 + \frac{1}{x^4} \cdot x^4 = x^4 + 1 \Rightarrow \text{St...
Let’s try plugging values:
Let $ x = \frac{5 + \sqrt{21}}{2} $, then $ x + \frac{1}{x} = 5 $. Com...
x = \frac{5 + \sqrt{21}}{2} \approx \frac{5 + 4.583}{2} = \frac{9.583}...
Then $ x^2 \approx 22.96 $, so:

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