# 4-x-x-1-5-4-6x-x-1-5-1-x-1-x-2-

Question Number 11515 by @ANTARES_VY last updated on 27/Mar/17
$$\sqrt[{\mathrm{5}}]{\frac{\mathrm{4}+\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}}}−\sqrt[{\mathrm{5}}]{\frac{\mathrm{4}−\mathrm{6}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}}}=\mathrm{1}. \\$$$$\boldsymbol{\mathrm{x}}_{\mathrm{1}} +\boldsymbol{\mathrm{x}}_{\mathrm{2}} =? \\$$
Answered by ajfour last updated on 27/Mar/17
$$\mathrm{x}=\frac{\mathrm{4}}{\mathrm{15}} \\$$
Commented by ajfour last updated on 27/Mar/17
$$\sqrt{\frac{\mathrm{4}+\mathrm{x}}{\mathrm{x}}}+\sqrt{\frac{\mathrm{4}−\mathrm{6x}}{\mathrm{x}}}=\mathrm{1}\:\:\:….\left(\mathrm{i}\right) \\$$$$\mathrm{let}\:\:\sqrt{\frac{\mathrm{4}+\mathrm{x}}{\mathrm{x}}}−\sqrt{\frac{\mathrm{4}−\mathrm{6x}}{\mathrm{x}}}=\mathrm{a}\:\:\:\:…\left(\mathrm{ii}\right) \\$$$$\mathrm{multiplying}\:\mathrm{above}\:\mathrm{two}\:\mathrm{equations} \\$$$$\left(\frac{\mathrm{4}+\mathrm{x}}{\mathrm{x}}\:\right)−\:\left(\frac{\mathrm{4}−\mathrm{6x}}{\mathrm{x}}\:\right)=\mathrm{a} \\$$$$\Rightarrow\:\:\mathrm{7}=\mathrm{a} \\$$$$\mathrm{adding}\:\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right): \\$$$$\mathrm{2}\sqrt{\frac{\mathrm{4}+\mathrm{x}}{\mathrm{x}}}\:=\:\mathrm{a}+\mathrm{1}\:=\:\mathrm{8}\:\:\left(\mathrm{as}\:\mathrm{a}=\mathrm{7}\:\right) \\$$$$\Rightarrow\:\:\frac{\mathrm{4}+\mathrm{x}}{\mathrm{x}}\:=\mathrm{16} \\$$$$\Rightarrow\:\mathrm{15x}\:=\mathrm{4} \\$$$$\:\:\:\mathrm{x}\:=\:\frac{\mathrm{4}}{\mathrm{15}}\:\:. \\$$
Commented by mrW1 last updated on 27/Mar/17
$${in}\:{the}\:{equation}\:{it}\:{is}\:{not}\:\sqrt{\:\:\:},\:{but}\:\:^{\mathrm{5}} \sqrt{\:\:}. \\$$
Commented by @ANTARES_VY last updated on 27/Mar/17
$$\boldsymbol{\mathrm{understand}}\:\:\boldsymbol{\mathrm{solved}} \\$$
Commented by chux last updated on 27/Mar/17
$$\mathrm{please}\:\mathrm{which}\:\mathrm{keyboard}\:\mathrm{editor}\:\mathrm{do}\: \\$$$$\mathrm{you}\:\mathrm{use}? \\$$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Mar/17
$$\underset{} {{l}et}:\frac{\mathrm{4}+{x}}{{x}}={a}^{\mathrm{5}} \Rightarrow\frac{\mathrm{4}}{{x}}+\mathrm{1}={a}^{\mathrm{5}} \Rightarrow\frac{\mathrm{4}}{{x}}−\mathrm{6}={a}^{\mathrm{5}} −\mathrm{7} \\$$$${a}−\sqrt[{\mathrm{5}}]{{a}^{\mathrm{5}} −\mathrm{7}}=\mathrm{1}\Rightarrow\left({a}−\mathrm{1}\right)^{\mathrm{5}} ={a}^{\mathrm{5}} −\mathrm{7} \\$$$${a}^{\mathrm{5}} −\mathrm{5}{a}^{\mathrm{4}} +\mathrm{10}{a}^{\mathrm{3}} −\mathrm{10}{a}^{\mathrm{2}} +\mathrm{5}{a}−\mathrm{1}={a}^{\mathrm{5}} −\mathrm{7} \\$$$$\mathrm{5}{a}^{\mathrm{4}} −\mathrm{10}{a}^{\mathrm{3}} +\mathrm{10}{a}^{\mathrm{2}} −\mathrm{5}{a}−\mathrm{6}=\mathrm{0} \\$$$$\Rightarrow{a}=\mathrm{1}.\mathrm{47}\Rightarrow\frac{\mathrm{4}+{x}}{{x}}=\left(\mathrm{1}.\mathrm{47}\right)^{\mathrm{5}} =\mathrm{6}.\mathrm{86} \\$$$$\frac{\mathrm{4}}{{x}}=\mathrm{5}.\mathrm{86}\Rightarrow{x}=\frac{\mathrm{4}}{\mathrm{5}.\mathrm{86}}=.\mathrm{682}\:\:. \\$$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Mar/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Mar/17