Question Number 224873 by ajfour last updated on 08/Oct/25
Commented by ajfour last updated on 08/Oct/25
$${If}\:\:{x}^{\mathrm{2}} +{bx}=−{c} \\ $$
Commented by ajfour last updated on 08/Oct/25
Commented by ajfour last updated on 08/Oct/25
$${because} \\ $$$${s}\mathrm{cos}\:\theta−{s}\mathrm{sin}\:\theta={b} \\ $$$${s}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta=−{c} \\ $$$${s}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2sin}\:\theta\mathrm{cos}\:\theta\right)={b}^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}{c}}{{s}^{\mathrm{2}} }\right)={b}^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} ={b}^{\mathrm{2}} −\mathrm{2}{c} \\ $$$${say}\:\mathrm{sin}\:\theta={q} \\ $$$${q}^{\mathrm{2}} \left(\mathrm{1}−{q}^{\mathrm{2}} \right)=\frac{{c}^{\mathrm{2}} }{{s}^{\mathrm{4}} } \\ $$$${q}^{\mathrm{4}} −{q}^{\mathrm{2}} +\frac{{c}^{\mathrm{2}} }{{s}^{\mathrm{4}} }=\mathrm{0} \\ $$$${q}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{{c}^{\mathrm{2}} }{{s}^{\mathrm{4}} }} \\ $$$${x}^{\mathrm{2}} ={s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${x}^{\mathrm{2}} =\left({b}^{\mathrm{2}} −\mathrm{2}{c}\right)\left\{\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left(\frac{{c}}{{b}^{\mathrm{2}} −\mathrm{2}{c}}\right)^{\mathrm{2}} }\right\} \\ $$$$ \\ $$
Commented by fantastic last updated on 08/Oct/25
$${Q}\mathrm{224859}.\:{sir}\:{can}\:{you}\:{please}\:{help}\:{me} \\ $$
Commented by ajfour last updated on 09/Oct/25
https://youtu.be/F6bd8sNM-3c?si=Zv8zgNANG0uy6WHE
Commented by mr W last updated on 13/Oct/25
$${please}\:{give}\:{a}\:{try}\:{to}\:\:{Q}\mathrm{224853} \\ $$