Question Number 116509 by Khalmohmmad last updated on 04/Oct/20
$$\mathrm{x}^{\mathrm{4}} −\mathrm{48x}^{\mathrm{2}} +\mathrm{x}+\mathrm{565}=\mathrm{0}\: \\ $$$$\mathrm{x}=? \\ $$
Answered by TANMAY PANACEA last updated on 04/Oct/20
$${f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{48}{x}^{\mathrm{2}} +\mathrm{576}+{x}−\mathrm{11} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{24}\right)^{\mathrm{2}} +{x}−\mathrm{11} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{576}−\mathrm{11}>\mathrm{0} \\ $$$${f}\left(\mathrm{4}\right)=\left(−\mathrm{8}\right)^{\mathrm{2}} −\mathrm{7}>\mathrm{0} \\ $$$$\boldsymbol{{f}}\left(\mathrm{5}\right)=\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}<\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{one}}\:\boldsymbol{{root}}\:\:\:\:\mathrm{5}>\boldsymbol{{x}}>\mathrm{4} \\ $$$$\boldsymbol{{wait}}… \\ $$$${f}\left(\mathrm{6}\right)=\left(\mathrm{36}−\mathrm{24}\right)^{\mathrm{2}} +\mathrm{6}−\mathrm{11}>\mathrm{0} \\ $$$${so}\:{another}\:{root}\:\mathrm{6}>{x}>\mathrm{5} \\ $$
Answered by MJS_new last updated on 04/Oct/20
$$\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{5}.\mathrm{29497} \\ $$$${x}_{\mathrm{2}} \approx−\mathrm{4}.\mathrm{47946} \\ $$$${x}_{\mathrm{3}} \approx\mathrm{4}.\mathrm{63433} \\ $$$${x}_{\mathrm{4}} \approx\mathrm{5}.\mathrm{14011} \\ $$