Elimination/Zwei quadratische Gleichungen/Direkt/Fakt/Beweis

Beweis

Wir haben zunächst

${}a_{2}F_{1}-a_{1}F_{2}=(a_{2}b_{1}-a_{1}b_{2})X+a_{2}c_{1}-a_{1}c_{2}\,.$

Wir ziehen das ${}a_{1}$-fache dieses Ausdrucks vom ${}(a_{2}b_{1}-a_{1}b_{2})$-fachen von ${}F_{1}$ ab und erhalten

$(b_{1}(a_{2}b_{1}-a_{1}b_{2})-a_{1}(a_{2}b_{1}-a_{1}b_{2}))X+(a_{2}b_{1}-a_{1}b_{2})c_{1}-a_{1}(a_{2}c_{1}-a_{1}c_{2}).$

Indem wir dies wieder über Kreuz multiplizieren erhalten wir

$(b_{1}(a_{2}b_{1}-a_{1}b_{2})-a_{1}(a_{2}b_{1}-a_{1}b_{2}))(a_{2}c_{1}-a_{1}c_{2})-(a_{2}b_{1}-a_{1}b_{2})((a_{2}b_{1}-a_{1}b_{2})c_{1}-a_{1}(a_{2}c_{1}-a_{1}c_{2})).$

$(a_{2}c_{1}-a_{1}c_{2})^{2}-b_{1}(-a_{2}c_{1}b_{2}-c_{2}a_{2}b_{1}+a_{1}b_{2}c_{2})+c_{1}(a_{1}b_{2}^{2}-2a_{2}b_{1}b_{2}).$

Zur bewiesenen Aussage