Question Number 10204 by gogel last updated on 30/Jan/17
$${x}^{\mathrm{2}} {p}^{\mathrm{2}} +{xyp}−\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by FilupSmith last updated on 30/Jan/17
$$\mathrm{solve}\:\mathrm{for}\:{x}: \\ $$$$\: \\ $$$${a}={p}^{\mathrm{2}} ,\:{b}={yp},\:{c}=−\mathrm{6}{y}^{\mathrm{2}} \\ $$$$\therefore\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\therefore\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\therefore\:{x}=\frac{−{yp}\pm\sqrt{{y}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{24}{p}^{\mathrm{2}} {y}^{\mathrm{2}} }}{\mathrm{2}{p}^{\mathrm{2}} } \\ $$$$\therefore\:{x}=\frac{−{yp}\pm\sqrt{\mathrm{25}{p}^{\mathrm{2}} {y}^{\mathrm{2}} }}{\mathrm{2}{p}^{\mathrm{2}} } \\ $$$$\therefore\:{x}=\frac{−{yp}\pm\mathrm{5}{py}}{\mathrm{2}{p}^{\mathrm{2}} } \\ $$$$\therefore\:{x}=\frac{\left(−\mathrm{1}\pm\mathrm{5}\right){y}}{\mathrm{2}{p}} \\ $$$${x}\:=\:−\frac{\mathrm{3}{y}}{{p}},\:\frac{\mathrm{2}{y}}{{p}}\: \\ $$