Question Number 91744 by ajfour last updated on 02/May/20
Commented by ajfour last updated on 02/May/20
$${If}\:\:{the}\:{red}\:{cubic}\:{curve}\:{has}\:{equation} \\ $$$$\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}}\:\:\:\:{while}\:{the} \\ $$$${blue}\:{one}\:{is}\:{the}\:{same}\:{red} \\ $$$${one},\:{shifted}\:{such}\:{that}\:{two}\:{roots} \\ $$$${are}\:{still}\:{common},\:{find}\:{equation} \\ $$$${of}\:{the}\:{blue}\:{one}\:{in}\:{terms}\:{of} \\ $$$$\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}.\boldsymbol{{x}},\:\boldsymbol{{y}}. \\ $$
Answered by mr W last updated on 02/May/20
![y=x^3 +ax^2 +bx+c with p=((3b−a^2 )/9) q=((2a^3 −9ab+27c)/(54)) and p^3 +q^2 <0 (such that three real roots) we get the three roots α=2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 )))−((2π)/3))−(a/3) β=2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 ))))−(a/3) γ=2(√(−p)) sin ((1/3)sin^(−1) (q/(√(−p^3 )))+((2π)/3))−(a/3) the eqn. of shifted cubic curve is y−k=(x−h)^3 +a(x−h)^2 +b(x−h)+c or y=(x−h)^3 +a(x−h)^2 +b(x−h)+c+k since β and γ are also its roots, (β−h)^3 +a(β−h)^2 +b(β−h)+c+k=0 ..(i) (γ−h)^3 +a(γ−h)^2 +b(γ−h)+c+k=0 ..(ii) (i)−(ii): 3h^2 −[2a+3(β+γ)]h+[β^2 +γ^2 +βγ+a(β+γ)+b]=0 ⇒h=((2a+3(β+γ)+(√(4a^2 −12b−3(β−γ)^2 )))/6) ⇒k=−(β−h)^3 −a(β−h)^2 −b(β−h)−c](Q91747.png)
$${y}={x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${with} \\ $$$${p}=\frac{\mathrm{3}{b}−{a}^{\mathrm{2}} }{\mathrm{9}} \\ $$$${q}=\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{54}} \\ $$$${and} \\ $$$${p}^{\mathrm{3}} +{q}^{\mathrm{2}} <\mathrm{0}\:\left({such}\:{that}\:{three}\:{real}\:{roots}\right) \\ $$$${we}\:{get}\:{the}\:{three}\:{roots} \\ $$$$\alpha=\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\frac{{a}}{\mathrm{3}} \\ $$$$\beta=\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}\right)−\frac{{a}}{\mathrm{3}} \\ $$$$\gamma=\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\sqrt{−{p}^{\mathrm{3}} }}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\frac{{a}}{\mathrm{3}} \\ $$$$ \\ $$$${the}\:{eqn}.\:{of}\:{shifted}\:{cubic}\:{curve}\:{is} \\ $$$${y}−{k}=\left({x}−{h}\right)^{\mathrm{3}} +{a}\left({x}−{h}\right)^{\mathrm{2}} +{b}\left({x}−{h}\right)+{c} \\ $$$${or} \\ $$$${y}=\left({x}−{h}\right)^{\mathrm{3}} +{a}\left({x}−{h}\right)^{\mathrm{2}} +{b}\left({x}−{h}\right)+{c}+{k} \\ $$$${since}\:\beta\:{and}\:\gamma\:{are}\:{also}\:{its}\:{roots}, \\ $$$$\left(\beta−{h}\right)^{\mathrm{3}} +{a}\left(\beta−{h}\right)^{\mathrm{2}} +{b}\left(\beta−{h}\right)+{c}+{k}=\mathrm{0}\:\:..\left({i}\right) \\ $$$$\left(\gamma−{h}\right)^{\mathrm{3}} +{a}\left(\gamma−{h}\right)^{\mathrm{2}} +{b}\left(\gamma−{h}\right)+{c}+{k}=\mathrm{0}\:\:..\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{3}{h}^{\mathrm{2}} −\left[\mathrm{2}{a}+\mathrm{3}\left(\beta+\gamma\right)\right]{h}+\left[\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\beta\gamma+{a}\left(\beta+\gamma\right)+{b}\right]=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{h}=\frac{\mathrm{2}{a}+\mathrm{3}\left(\beta+\gamma\right)+\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{12}{b}−\mathrm{3}\left(\beta−\gamma\right)^{\mathrm{2}} }}{\mathrm{6}} \\ $$$$\Rightarrow{k}=−\left(\beta−{h}\right)^{\mathrm{3}} −{a}\left(\beta−{h}\right)^{\mathrm{2}} −{b}\left(\beta−{h}\right)−{c} \\ $$
Commented by mr W last updated on 02/May/20
Commented by mr W last updated on 03/May/20
$${sir},\:{please}\:{also}\:{have}\:{a}\:{look}\:{at}\:{Q}\mathrm{91446}. \\ $$