Question Number 114388 by ajfour last updated on 18/Sep/20
![Developed the formula for the cubic equation when it has three real roots and Cardano is just not suitable as discriminant gets negative.. Given x^3 −bx−c=0 with b>0 , c>0 (t/b)=1+((15)/2)((c^2 /b^3 ))+(√(((c^2 /b^3 ))[49−((75)/4)((c^2 /b^3 ))])) x=(√(t+((4c^2 )/b^2 ))) −((3c)/b) ★ mrW Sir, MjS Sir please check!](Q114388.png)
$${Developed}\:{the}\:{formula}\:{for}\:{the}\:{cubic} \\ $$ $${equation}\:{when}\:{it}\:{has}\:{three}\:{real}\:{roots} \\ $$ $${and}\:{Cardano}\:{is}\:{just}\:{not}\:{suitable}\:{as} \\ $$ $${discriminant}\:{gets}\:{negative}.. \\ $$ $$\:\:\:{Given}\:\:\:\:\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{bx}}−\boldsymbol{{c}}=\mathrm{0} \\ $$ $$\:\:{with}\:\:\boldsymbol{{b}}>\mathrm{0}\:,\:\boldsymbol{{c}}>\mathrm{0} \\ $$ $$\:\frac{\boldsymbol{{t}}}{\boldsymbol{{b}}}=\mathrm{1}+\frac{\mathrm{15}}{\mathrm{2}}\left(\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{3}} }\right)+\sqrt{\left(\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{3}} }\right)\left[\mathrm{49}−\frac{\mathrm{75}}{\mathrm{4}}\left(\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{3}} }\right)\right]} \\ $$ $$\:\:\:\:\:\boldsymbol{{x}}=\sqrt{\boldsymbol{{t}}+\frac{\mathrm{4}\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }}\:−\frac{\mathrm{3}\boldsymbol{{c}}}{\boldsymbol{{b}}}\:\:\bigstar \\ $$ $${mrW}\:{Sir},\:{MjS}\:{Sir}\:\:{please}\:{check}! \\ $$
Commented byMJS_new last updated on 18/Sep/20
$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{exact} \\ $$ $${x}^{\mathrm{3}} −\frac{\mathrm{244}}{\mathrm{3}}{x}−\frac{\mathrm{1456}}{\mathrm{27}}=\mathrm{0}\:\Rightarrow\:{x}\in\left\{−\frac{\mathrm{26}}{\mathrm{3}},\:−\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{28}}{\mathrm{3}}\right\} \\ $$ $$\mathrm{but}\:\mathrm{your}\:\mathrm{formula}\:\mathrm{gives} \\ $$ $${x}\approx\mathrm{9}.\mathrm{33355157}\:\mathrm{instead}\:\mathrm{of}\:\mathrm{9}.\overset{\bullet} {\mathrm{3}} \\ $$
Commented byajfour last updated on 18/Sep/20
$${yes}\:{sir},\:{i}\:{found}\:{the}\:{error}!\:{thanks}. \\ $$