Pagina:Werkecarlf03gausrich.djvu/360

${\begin{aligned}\alpha \alpha ^{\prime }\alpha ^{\prime \prime }&={\phantom {-}}{\frac {\varepsilon aabAAB}{(G+G^{\prime })(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}\\\beta \beta ^{\prime }\beta ^{\prime \prime }&=-{\frac {\varepsilon abbABB}{(G+G^{\prime })(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}\\\alpha \beta &={\phantom {-}}{\frac {abAB}{(G+G^{\prime })(G+G^{\prime \prime })}}\\\alpha ^{\prime }\beta ^{\prime }&=-{\frac {abAB}{(G+G^{\prime })(G^{\prime }-G^{\prime \prime })}}\\\alpha ^{\prime \prime }\beta ^{\prime \prime }&={\phantom {-}}{\frac {abAB}{(G+G^{\prime \prime })(G^{\prime }-G^{\prime \prime })}}\end{aligned}}$

10.

Formulae nostrae quibusdam casibus indeterminatae fieri possunt, quos seorsim considerare oportet. Ac primo quidem discutiemus casum eum, ubi aequationis cubicae radices negativae $-G^{\prime },$ $-G^{\prime \prime }$ aequales fiunt, unde, per formulas 18, 19, coëfficientes $\gamma ,$ $\gamma ^{\prime \prime }$ valores infinitos nancisci videntur, qui autem revera sunt indeterminati.

Statuendo in formula 14, $g=G,$ patet, ut duo valores ipsius $x,$ i.e. ut $-G^{\prime }$ et $-G^{\prime \prime }$ fiant aequales, necessario esse debere

$AB=0,\quad aa-bb-{\frac {aaAA}{aa+G}}+{\frac {bbBB}{bb+G}}=0$

Hinc facile intelligitur, quum $aa-bb$ natura sua sit vel quantitas positiva, vel $=0,$ esse debere

$B=0$

$aa-bb={\frac {aaAA}{aa+G}},\quad$ sive $\quad aa+G={\frac {aaAA}{aa-bb}}$

Substituendo hos valores in aequatione 14, fit

$G^{\prime }=G^{\prime \prime }=bb$

Substituendo porro valorem $x=-bb$ in aequatione cubica 13, prodit

$(aa-bb)(CC+bb)=bbAA$

Quoties haec aequatio conditionalis simul cum aequatione $B=0$ locum habet, casus, quem hic tractamus, adducitur. Et quum fiat

$G={\frac {aaAA}{aa-bb}}-aa={\frac {aaCC}{bb}}$