Appendix A1, PDF p. 28 (printed p. 22) — r₁ − t = −t(e + 4t − √(3(4t² − e²)))/(2(e + t)), and the numerator is positive

Initializing...

Appendix A1, PDF p. 28 (printed p. 22) — r₁ − t = −t(e + 4t − √(3(4t² − e²)))/(2(e + t)), and the numerator is positive · Prove2Me