Question Number 69297 by ajfour last updated on 22/Sep/19
Commented by ajfour last updated on 22/Sep/19
$${If}\:{the}\:{graph}\:{is}\:{a}\:{cubic} \\ $$$$\:\:\:\:{y}={x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${find}\:{point}\:{of}\:{intersection}\:{of} \\ $$$${the}\:{red}\:{and}\:{blue}\:{lines}. \\ $$
Answered by MJS last updated on 22/Sep/19
![y=x^3 +ax^2 +bx+c if we shift this so that the middle zero is at x=0 we get y=(x+α^2 )x(x−β^2 ) then it′s easy to find the tangents because the tangent in P is y=(3p^2 +2(α^2 −β^2 )p−α^2 β^2 )x−(2p+α^2 −β^2 )p^2 putting x=−α^2 or x=β^2 we know we′ll get a double solution at −α^2 or β^2 ⇒ the points on the curve P_(−α^2 ) = (((β^2 /2)),((−((2α^2 +β^2 )/8)β^4 )) ) P_β^2 = (((−(α^2 /2))),((((α^2 +2β^2 )/8)α^4 )) ) the tangents: y_(−α^2 ) =−(β^4 /4)x−((α^2 β^4 )/4) y_β^2 =−(α^4 /4)x+((α^4 β^2 )/4) the intersection is I= ((((α^2 β^2 )/(α^2 −β^2 ))),((−((α^4 β^4 )/(4(α^2 −β^2 ))))) ) we have to find α^2 , β^2 in terms of a, b, c which is possible but nasty because we have to use the trigonometric method... x^3 +ax^2 +bx+c=0 x_1 <x_2 <x_3 x=t+x_2 ⇒ t^3 +(a+3x_2 )t^2 +(3x_2 ^2 +2ax_2 +b)t=0 [because the constant factor x_2 ^3 +ax_2 ^3 +bx_2 +c is equal to zero] ⇒ α^2 =−((3x_2 +a)/2)−((√(−3x_2 ^2 −2ax_2 +a^2 −4b))/2) β^2 =−((3x_2 +a)/2)+((√(−3x_2 ^2 −2ax_2 +a^2 −4b))/2) x_2 =−(a/3)+(2/3)(√(a^2 −3b))sin (((2π)/3)k+(1/3)arcsin ((2a^3 −9ab+27c)/(2(√((a^2 −3b)^3 ))))) with k=0, 1, 2 I′m not sure but I think the middle zero is the one with z=0](Q69315.png)
$${y}={x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{shift}\:\mathrm{this}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{zero}\:\mathrm{is}\:\mathrm{at} \\ $$$${x}=\mathrm{0}\:\mathrm{we}\:\mathrm{get} \\ $$$${y}=\left({x}+\alpha^{\mathrm{2}} \right){x}\left({x}−\beta^{\mathrm{2}} \right) \\ $$$$\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{because} \\ $$$$\mathrm{the}\:\mathrm{tangent}\:\mathrm{in}\:{P}\:\mathrm{is} \\ $$$${y}=\left(\mathrm{3}{p}^{\mathrm{2}} +\mathrm{2}\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right){p}−\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \right){x}−\left(\mathrm{2}{p}+\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right){p}^{\mathrm{2}} \\ $$$$\mathrm{putting}\:{x}=−\alpha^{\mathrm{2}} \:\mathrm{or}\:{x}=\beta^{\mathrm{2}} \:\mathrm{we}\:\mathrm{know}\:\mathrm{we}'\mathrm{ll}\:\mathrm{get} \\ $$$$\mathrm{a}\:\mathrm{double}\:\mathrm{solution}\:\mathrm{at}\:−\alpha^{\mathrm{2}} \:\mathrm{or}\:\beta^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve} \\ $$$${P}_{−\alpha^{\mathrm{2}} } =\begin{pmatrix}{\frac{\beta^{\mathrm{2}} }{\mathrm{2}}}\\{−\frac{\mathrm{2}\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\mathrm{8}}\beta^{\mathrm{4}} }\end{pmatrix}\:\:\:{P}_{\beta^{\mathrm{2}} } =\begin{pmatrix}{−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}}\\{\frac{\alpha^{\mathrm{2}} +\mathrm{2}\beta^{\mathrm{2}} }{\mathrm{8}}\alpha^{\mathrm{4}} }\end{pmatrix} \\ $$$$\mathrm{the}\:\mathrm{tangents}: \\ $$$${y}_{−\alpha^{\mathrm{2}} } =−\frac{\beta^{\mathrm{4}} }{\mathrm{4}}{x}−\frac{\alpha^{\mathrm{2}} \beta^{\mathrm{4}} }{\mathrm{4}} \\ $$$${y}_{\beta^{\mathrm{2}} } =−\frac{\alpha^{\mathrm{4}} }{\mathrm{4}}{x}+\frac{\alpha^{\mathrm{4}} \beta^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{the}\:\mathrm{intersection}\:\mathrm{is} \\ $$$${I}=\begin{pmatrix}{\frac{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} }{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }}\\{−\frac{\alpha^{\mathrm{4}} \beta^{\mathrm{4}} }{\mathrm{4}\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)}}\end{pmatrix} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a},\:{b},\:{c}\:\mathrm{which} \\ $$$$\mathrm{is}\:\mathrm{possible}\:\mathrm{but}\:\mathrm{nasty}\:\mathrm{because}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{use} \\ $$$$\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{method}... \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} <{x}_{\mathrm{2}} <{x}_{\mathrm{3}} \\ $$$${x}={t}+{x}_{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{3}} +\left({a}+\mathrm{3}{x}_{\mathrm{2}} \right){t}^{\mathrm{2}} +\left(\mathrm{3}{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{ax}_{\mathrm{2}} +{b}\right){t}=\mathrm{0} \\ $$$$\left[\mathrm{because}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{factor}\right. \\ $$$$\left.\:\:{x}_{\mathrm{2}} ^{\mathrm{3}} +{ax}_{\mathrm{2}} ^{\mathrm{3}} +{bx}_{\mathrm{2}} +{c}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{zero}\right] \\ $$$$\Rightarrow \\ $$$$\alpha^{\mathrm{2}} =−\frac{\mathrm{3}{x}_{\mathrm{2}} +{a}}{\mathrm{2}}−\frac{\sqrt{−\mathrm{3}{x}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{ax}_{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}} \\ $$$$\beta^{\mathrm{2}} =−\frac{\mathrm{3}{x}_{\mathrm{2}} +{a}}{\mathrm{2}}+\frac{\sqrt{−\mathrm{3}{x}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{ax}_{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =−\frac{{a}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{a}^{\mathrm{2}} −\mathrm{3}{b}}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}{k}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{2}\sqrt{\left({a}^{\mathrm{2}} −\mathrm{3}{b}\right)^{\mathrm{3}} }}\right) \\ $$$$\:\:\:\:\:\mathrm{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{zero}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{one}\:\mathrm{with}\:{z}=\mathrm{0} \\ $$
Commented by ajfour last updated on 22/Sep/19
$${thanks}\:{sir},\:{i}\:{expect}\:{more}\:{out}\:{of} \\ $$$${this}\:{construction}..{say}\:{extract} \\ $$$${roots}. \\ $$