1. if F = f = 0,
then $D = EG, H = H = \frac{eG - 2fF + gE}{D}=\frac{eG+gE}{2D}, K = K = \frac{eg -f^{2}}{D}= \frac{eg}{D}$,
$k_{1}, k_{2} = H \pm  \sqrt{H^{2}+K} = \frac{eG+gE}{2D} \pm \sqrt{(\frac{eG+gE}{2D})^{2} - \frac{eg}{D}} = \frac{eG+gE}{2EG} \pm \sqrt{(\frac{eG+gE}{2EG})^{2} - \frac{eg}{EG}} = \frac{e}{2E} + \frac{g}{2G}  \pm \sqrt{\frac{e^{2}G^{2} +2EeGg+g^{2}E^{2}}{4E^{2}G^{2}} - \frac{eg}{EG}} = \frac{e}{2E} + \frac{g}{2G}  \pm \sqrt{\frac{e^{2}G^{2} +2EeGg+g^{2}E^{2} -4EeGg}{4E^{2}G^{2}}} = \frac{e}{2E} + \frac{g}{2G}  \pm \sqrt{\frac{e^{2}G^{2} -2EeGg+g^{2}E^{2}}{4E^{2}G^{2}}} = \frac{e}{2E} + \frac{g}{2G}  \pm \sqrt{(\frac{eG -gE}{2EG})^{2}} = \frac{e}{2E} + \frac{g}{2G}  \pm (\frac{eG -gE}{2EG})$
$k_{1} = \frac{e}{2E} + \frac{g}{2G} + \frac{eG -gE}{2EG} = \frac{e}{2E} + \frac{g}{2G} +\frac{e}{2E} - \frac{g}{2G} = \frac{e}{E}$
$k_{2} = \frac{e}{2E} + \frac{g}{2G} - \frac{eG -gE}{2EG} = \frac{e}{2E} + \frac{g}{2G} -\frac{e}{2E} + \frac{g}{2G} = \frac{g}{G}$