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Question-187059

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Question Number 187059 by ajfour last updated on 13/Feb/23
Commented by ajfour last updated on 13/Feb/23
Bigger curve y=x^3 −x−c the other y=m(x−(1/( (√3))))^2 (x−s) say their intersections be at x=p, q, t Also p^3 =p+c we must find p in surds...even if 0<c<(2/(3(√3))) ★
$${Bigger}\:{curve}\:\:\:{y}={x}^{\mathrm{3}} −{x}−{c} \\ $$$${the}\:{other}\:\:\:{y}={m}\left({x}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \left({x}−{s}\right) \\ $$$${say}\:{their}\:{intersections}\:{be}\:{at} \\ $$$${x}={p},\:{q},\:{t} \\ $$$${Also}\:\:{p}^{\mathrm{3}} ={p}+{c} \\ $$$${we}\:{must}\:{find}\:{p}\:{in}\:{surds}…{even}\:{if} \\ $$$$\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\bigstar \\ $$
Answered by ajfour last updated on 13/Feb/23
y=x^3 −x Y=m(x−(1/( (√3))))^2 (x−s) =m(x^2 −((2x)/( (√3)))+(1/3))(x−s) y=Y=c=p^3 −p =m{p^3 −(s+(2/( (√3))))p^2 +(((2s)/( (√3)))+(1/3))p−(s/3)} ⇒(m−1)p^3 −m(s+(2/( (√3))))p^2 +{m(((2s)/( (√3)))+(1/3))+1}p−((ms)/3)=0 let m^2 (s+(2/( (√3))))^2 =3m(m−1)(((2s)/( (√3)))+(1/3)) +3(m−1) let s=−(1/(2(√3))) ⇒ m^2 =4(m−1) ⇒ (m−2)^2 =0 ⇒ m=2 Now c=2{p^3 −(((√3)p^2 )/2)+(1/(6(√3)))} and for c=(1/3) p^3 −(((√3)p^2 )/2)−(1/6)(1−(1/( (√3))))=0 let me check this...
$${y}={x}^{\mathrm{3}} −{x} \\ $$$${Y}={m}\left({x}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \left({x}−{s}\right) \\ $$$$={m}\left({x}^{\mathrm{2}} −\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({x}−{s}\right) \\ $$$${y}={Y}={c}={p}^{\mathrm{3}} −{p} \\ $$$$={m}\left\{{p}^{\mathrm{3}} −\left({s}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right){p}^{\mathrm{2}} +\left(\frac{\mathrm{2}{s}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{3}}\right){p}−\frac{{s}}{\mathrm{3}}\right\} \\ $$$$\Rightarrow\left({m}−\mathrm{1}\right){p}^{\mathrm{3}} −{m}\left({s}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right){p}^{\mathrm{2}} \\ $$$$\:\:+\left\{{m}\left(\frac{\mathrm{2}{s}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{1}\right\}{p}−\frac{{ms}}{\mathrm{3}}=\mathrm{0} \\ $$$${let}\:\: \\ $$$${m}^{\mathrm{2}} \left({s}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\mathrm{3}{m}\left({m}−\mathrm{1}\right)\left(\frac{\mathrm{2}{s}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}\left({m}−\mathrm{1}\right) \\ $$$${let}\:\:{s}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\:{m}^{\mathrm{2}} =\mathrm{4}\left({m}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\left({m}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0}\:\:\Rightarrow\:\:{m}=\mathrm{2} \\ $$$${Now} \\ $$$${c}=\mathrm{2}\left\{{p}^{\mathrm{3}} −\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}\sqrt{\mathrm{3}}}\right\} \\ $$$${and}\:{for}\:\:{c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${p}^{\mathrm{3}} −\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{0} \\ $$$${let}\:{me}\:{check}\:{this}… \\ $$

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