Question Number 27046 by ajfour last updated on 01/Jan/18
![Considering y=x^3 +px+q If (dy/dx)∣_(x=α) =0 ⇒ α^2 =−(p/3) if ((d(y/x))/dx)∣_(x=β) =0 ⇒ β^( 3) =(q/2) roots of the cubic eq^n are: x=[−β^( 3) ±(√(β^( 6) −α^6 )) ]^(1/3) −[β^( 3) ±(√(β^( 6) −α^6 )) ]^(1/3) . Why such a connection? If equation is quadratic even_ y=ax^2 +bx+c (dy/dx)∣_(x=α) =0 ⇒ α=−(b/(2a)) ((d(y/x))/dx)∣_(x=β) =0 ⇒ β^( 2) =(c/a) roots of quadratic eq. are: x=𝛂±(√(𝛂^2 −𝛃^( 2) )) why such a connection ?](https://www.tinkutara.com/question/Q27046.png)
$${Considering}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}} \\ $$$${If}\:\:\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\Rightarrow\:\:\alpha^{\mathrm{2}} =−\frac{{p}}{\mathrm{3}} \\ $$$${if}\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\:\Rightarrow\:\beta^{\:\mathrm{3}} =\frac{{q}}{\mathrm{2}} \\ $$$${roots}\:{of}\:{the}\:{cubic}\:\:{eq}^{{n}} \:{are}: \\ $$$$\:\:\:\:{x}=\left[−\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left[\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \:. \\ $$$$\:{Why}\:{such}\:{a}\:{connection}? \\ $$$${If}\:{equation}\:{is}\:{quadratic}\:{even\_} \\ $$$$\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}} \\ $$$$\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\:\Rightarrow\:\:\alpha=−\frac{{b}}{\mathrm{2}{a}} \\ $$$$\:\:\:\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\Rightarrow\:\beta^{\:\mathrm{2}} =\frac{{c}}{{a}} \\ $$$${roots}\:{of}\:{quadratic}\:{eq}.\:{are}: \\ $$$$\:\:\:\:{x}=\boldsymbol{\alpha}\pm\sqrt{\boldsymbol{\alpha}^{\mathrm{2}} −\boldsymbol{\beta}^{\:\mathrm{2}} }\: \\ $$$${why}\:{such}\:{a}\:{connection}\:?\: \\ $$