Question Number 192780 by ajfour last updated on 27/May/23
Commented by ajfour last updated on 27/May/23
Find m in terms of c, hence p.
Answered by ajfour last updated on 27/May/23
$${y}={x}^{\mathrm{3}} −{x}−{c} \\ $$$$\mathrm{0}={p}^{\mathrm{3}} −{p}−{c} \\ $$$${y}={mx} \\ $$$${y}=−{x}^{\mathrm{2}} +{hx}+\mathrm{1}−{h} \\ $$$${c}=−{p}^{\mathrm{2}} +{hp}+\mathrm{1}−{h} \\ $$$${p}^{\mathrm{3}} ={p}+{c} \\ $$$${p}+{c}={h}\left({hp}+\mathrm{1}−{h}−{c}\right)+\left(\mathrm{1}−{h}−{c}\right){p} \\ $$$$\Rightarrow\:\:\left\{{h}^{\mathrm{2}} −\left({h}+{c}\right)\right\}{p}={h}^{\mathrm{2}} −\left(\mathrm{1}−{c}\right){h}+{c} \\ $$$${x}^{\mathrm{2}} +\left({m}−{h}\right){x}=\mathrm{1}−{h} \\ $$$${double}\:{root}\:\:\Rightarrow \\ $$$$\:\:\:\left({m}−{h}\right)^{\mathrm{2}} =\mathrm{4}\left({h}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:{h}^{\mathrm{2}} −\mathrm{4}{h}\left({m}+\mathrm{1}\right)+{m}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\left({h}−\mathrm{2}{m}−\mathrm{2}\right)^{\mathrm{2}} ={m}\left(\mathrm{3}{m}+\mathrm{8}\right) \\ $$$$\left\{\frac{{h}^{\mathrm{2}} −\left(\mathrm{1}−{c}\right){h}+{c}}{{h}^{\mathrm{2}} −{h}−{c}}\right\}^{\mathrm{2}} ={h}\left\{\frac{{h}^{\mathrm{2}} −\left(\mathrm{1}−{c}\right){h}+{c}}{{h}^{\mathrm{2}} −{h}−{c}}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{1}−{h}−{c} \\ $$$$\Rightarrow\:\left\{\frac{\left(\mathrm{4}{m}+\mathrm{3}+{c}\right){h}+{c}}{\left(\mathrm{4}{m}+\mathrm{3}\right){h}−{c}}\right\}^{\mathrm{2}} \\ $$$$\:\:\:={h}\left\{\frac{\left(\mathrm{4}{m}+\mathrm{3}+{c}\right){h}+{c}}{\left(\mathrm{4}{m}+\mathrm{3}\right){h}−{c}}\right\}+\mathrm{1}−{h}−{c} \\ $$$$…. \\ $$