Question Number 35892 by behi83417@gmail.com last updated on 25/May/18
Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18
$${i}\:{am}\:{very}\:{near}\:{to}\:{find}\:{the}\:{answer}…{its}\:{a}\:{very}\:{good} \\ $$$${problem}… \\ $$
Commented by Rasheed.Sindhi last updated on 29/May/18
$$\mathrm{We}'\mathrm{re}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{sir}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18
$${t}=\frac{−\mathrm{1}}{{y}}\:….\left(\mathrm{1}\right) \\ $$$${putting}\:{value}\:{of}\:{t}\:{in}\left({t}+{y}+{xz}=\mathrm{0}\right) \\ $$$${y}+{xz}−\frac{\mathrm{1}}{{y}}=\mathrm{0}\:\:\:…\left(\mathrm{2}\right) \\ $$$${x}+{z}=\mathrm{10}….\left(\mathrm{3}\right) \\ $$$${putting}\:{value}\:{in}\:\left({xt}+{yx}=\mathrm{6}\right) \\ $$$${x}.\frac{−\mathrm{1}}{{y}}+{yz}=\mathrm{6}….\left(\mathrm{4}\right) \\ $$$${putting}\:{z}=\mathrm{10}−{x}\:{in}\left({y}+{xz}−\frac{\mathrm{1}}{{y}}=\mathrm{0}\right) \\ $$$${y}+{x}\left(\mathrm{10}−{x}\right)−\frac{\mathrm{1}}{{y}}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +\mathrm{10}{xy}−{x}^{\mathrm{2}} {y}−\mathrm{1}=\mathrm{0} \\ $$$${putting}\:\:{z}=\mathrm{10}−{x}\:{in}\:\left(\frac{−{x}}{{y}}+{yz}=\mathrm{6}\right) \\ $$$$\frac{−{x}}{{y}}+{y}\left(\mathrm{10}−{x}\right)=\mathrm{6} \\ $$$$−{x}+\mathrm{10}{y}^{\mathrm{2}} −{xy}^{\mathrm{2}} =\mathrm{6}{y} \\ $$$${x}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)=\mathrm{10}{y}^{\mathrm{2}} −\mathrm{6}{y} \\ $$$${x}=\frac{\mathrm{10}{y}^{\mathrm{2}} −\mathrm{6}{y}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} +\mathrm{10}{xy}−{x}^{\mathrm{2}} {y}−\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +\mathrm{10}{y}\left(\frac{\mathrm{10}{y}^{\mathrm{2}} −\mathrm{6}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)=\mathrm{1}+{y}\left(\frac{\mathrm{10}{y}^{\mathrm{2}} −\mathrm{6}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\frac{{y}^{\mathrm{2}} +{y}^{\mathrm{4}} +\mathrm{100}{y}^{\mathrm{3}} −\mathrm{60}{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }=\mathrm{1}+{y}\frac{\left(\mathrm{100}{y}^{\mathrm{4}} −\mathrm{120}{y}^{\mathrm{3}} +\mathrm{36}{y}^{\mathrm{2}} \right.}{\mathrm{1}+\mathrm{2}{y}^{\mathrm{2}} +{y}^{\mathrm{4}} } \\ $$$${ditto}\:{lhs}=\frac{\mathrm{1}+\mathrm{2}{y}^{\mathrm{2}} +{y}^{\mathrm{4}} +\mathrm{100}{y}^{\mathrm{5}} −\mathrm{120}{y}^{\mathrm{4}} +\mathrm{36}{y}^{\mathrm{3}} }{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}+{y}_{} ^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} +{y}^{\mathrm{4}} +\mathrm{100}{y}^{\mathrm{3}} −\mathrm{60}{y}^{\mathrm{2}} \right)= \\ $$$$\mathrm{1}+\mathrm{2}{y}^{\mathrm{2}} +{y}^{\mathrm{4}} +\mathrm{100}{y}^{\mathrm{5}} −\mathrm{120}{y}^{\mathrm{4}} +\mathrm{36}{y}^{\mathrm{3}} \\ $$$${y}^{\mathrm{2}} +{y}^{\mathrm{4}} +\mathrm{100}{y}^{\mathrm{3}} −\mathrm{60}{y}^{\mathrm{2}} +{y}^{\mathrm{4}} +{y}^{\mathrm{6}} +\mathrm{100}{y}^{\mathrm{5}} −\mathrm{60}{y}^{\mathrm{4}} = \\ $$$$\mathrm{1}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{36}{y}^{\mathrm{3}} −\mathrm{119}{y}^{\mathrm{4}} +\mathrm{100}{y}^{\mathrm{5}} ={y}^{\mathrm{6}} +\mathrm{100}{y}^{\mathrm{5}} −\mathrm{58}{y}^{\mathrm{4}} \\ $$$$+\mathrm{100}{y}^{\mathrm{3}} −\mathrm{59}{y}^{\mathrm{2}} \\ $$$${contd} \\ $$$${y}^{\mathrm{6}} −\mathrm{58}{y}^{\mathrm{4}} +\mathrm{119}{y}^{\mathrm{4}} +\mathrm{100}{y}^{\mathrm{3}} −\mathrm{36}{y}^{\mathrm{3}} −\mathrm{59}{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{6}} +\mathrm{61}{y}^{\mathrm{4}} +\mathrm{64}{y}^{\mathrm{3}} −\mathrm{61}{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} +\mathrm{61}{y}+\mathrm{64}−\frac{\mathrm{61}}{{y}}−\frac{\mathrm{1}}{{y}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\left({y}^{\mathrm{3}} −\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\right)+\mathrm{61}\left({y}−\frac{\mathrm{1}}{{y}}\right)+\mathrm{64}=\mathrm{0} \\ $$$${k}={y}−\frac{\mathrm{1}}{{y}}\:\: \\ $$$${k}^{\mathrm{3}} ={y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} .\frac{\mathrm{1}}{{y}}+\mathrm{3}{y}.\frac{\mathrm{1}}{{y}^{\mathrm{2}} }−\frac{\mathrm{1}}{{y}^{\mathrm{3}} } \\ $$$${k}^{\mathrm{3}} ={y}^{\mathrm{3}} −\frac{\mathrm{1}}{{y}^{\mathrm{3}} }−\mathrm{3}\left({y}−\frac{\mathrm{1}}{{y}}\right) \\ $$$${k}^{\mathrm{3}} +\mathrm{3}{k}={y}^{\mathrm{3}} −\frac{\mathrm{1}}{{y}^{\mathrm{3}} } \\ $$$$\left({k}^{\mathrm{3}} +\mathrm{3}{k}\right)+\mathrm{61}{k}+\mathrm{64}=\mathrm{0} \\ $$$${k}^{\mathrm{3}} +\mathrm{64}{k}+\mathrm{64}=\mathrm{0} \\ $$$${k}\left({k}^{\mathrm{2}} +\mathrm{64}\right)+\mathrm{64}=\mathrm{0} \\ $$$${i}\:{have}\:{something}\:{to}\:{say}\:{about}\:{solvability}\:{of}\:{the} \\ $$$${eqution} \\ $$$$\left.\mathrm{1}\right){k}\:{should}\:{have}\:\left(−{ve}\right)\:{value} \\ $$$$\left.\mathrm{2}\right){but}\:{k}^{\mathrm{2}} +\mathrm{64}\:{is}\:+{ve}\:{so}\:{k}^{\mathrm{2}} +\mathrm{64}>\mathrm{64} \\ $$$$\left.\mathrm{3}\right){so}\:{k}\left({k}^{\mathrm{2}} +\mathrm{64}\right)\:{can}\:{never}\:{be}\:{equal}\:{to}\:\left(−\mathrm{64}\right) \\ $$$${to}\:{make}\:{the}\:{equation}\:{feasible} \\ $$$${so}\:{clnclusion}\:{k}^{\mathrm{3}} +\mathrm{64}{k}+\mathrm{64}\:{can}\:{not}\:{be}\:{zero} \\ $$$${i}\:{have}\:{to}\:{try}\:{to}\:{solve}\:{another}\:{way}.. \\ $$$${contd} \\ $$
Commented by behi83417@gmail.com last updated on 26/May/18
$${thank}\:{you}\:{very}\:{much}\:{sir}\:{for}\:{your}\:{working}. \\ $$$${i}\:{am}\:{waiting}\:{for}\:{your}\:{final}\:{answer}. \\ $$
Commented by behi83417@gmail.com last updated on 28/May/18
$${dear}\:{mr}\:{tanmay}!\:{i}\:{think}\:{one}\:{factor}\:{of} \\ $$$${your}\:{equation}\:{is}:\:{y}^{\mathrm{2}} +{y}−\mathrm{1} \\ $$$${and}\:{equation}:\:{k}^{\mathrm{3}} +\mathrm{64}{k}+\mathrm{64}=\mathrm{0}\:{at}\:{least} \\ $$$${have}\:{on}\:{root}\:{that}\:{apperoximally}\:{equails} \\ $$$${to}\::−\mathrm{1}\:\left({k}=−.\mathrm{9824}\right) \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/May/18
$${thanx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18
$${let}\:{y}={a}\:\:\:{t}=\frac{−\mathrm{1}}{{a}} \\ $$$${t}+{y}={a}−\frac{\mathrm{1}}{{a}} \\ $$$${t}+{y}=\frac{\left({a}+\mathrm{1}\right)\left({a}−\mathrm{1}\right)}{{a}} \\ $$$${so}\:{xz}=−\frac{\left({a}+\mathrm{1}\right)\left({a}−\mathrm{1}\right)}{{a}} \\ $$$${since}\:{t}+{y}+{xz}=\mathrm{0} \\ $$$${xz}=\frac{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}−{a}\right)}{{a}} \\ $$$${let} \\ $$$${x}=\frac{\mathrm{1}−{a}}{{a}}\:\:\:\:{z}=\mathrm{1}+{a} \\ $$$${x}=\frac{\mathrm{1}+{a}}{{a}}\:\:\:{z}=\mathrm{1}−{a} \\ $$$${x}=\mathrm{1}−{a}\:\:{z}=\frac{\mathrm{1}+{a}}{{a}} \\ $$$${x}+{z}=\mathrm{10} \\ $$$$\mathrm{1}+{a}+\frac{\mathrm{1}−{a}}{{a}}=\mathrm{10} \\ $$$${a}+{a}^{\mathrm{2}} +\mathrm{1}−{a}=\mathrm{10}{a} \\ $$$${a}^{\mathrm{2}} −\mathrm{10}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{4}}\:\:}{\mathrm{2}} \\ $$$${a}=\frac{\mathrm{10}\pm\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${a}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:\:\:{and}\:\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\:\: \\ $$$$\:{pls}\:{wait}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$