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The trace is invariant under Lorentz transformations, so: - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📝 Recent Posts

y(x^2 + 1) = 2x^2 - 3x + 1
yx^2 + y = 2x^2 - 3x + 1 \quad \Rightarrow \quad (y - 2)x^2 + 3x + (y...
For \(x\) to be real, the discriminant of this quadratic must be non-n...
3^2 - 4(y - 2)(y - 1) \geq 0
- 4(y^2 - 3y + 2) \geq 0
- 4y^2 + 12y - 8 \geq 0
-4y^2 + 12y + 1 \geq 0
Multiply through by \(-1\) (reversing the inequality):
4y^2 - 12y - 1 \leq 0
Solve the quadratic inequality using the quadratic formula:

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