Question Number 61343 by bhanukumarb2@gmail.com last updated on 01/Jun/19 | ||
Answered by MJS last updated on 01/Jun/19 | ||
$${x}={a}−\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}−...}} \\ $$$${x}={a}−\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} −{ax}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${a}=\mathrm{2207}\:\Rightarrow\:{x}=\frac{\mathrm{2207}\pm\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt[{\mathrm{8}}]{\frac{\mathrm{2207}\pm\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}}}=\sqrt{\sqrt{\sqrt{\frac{\mathrm{2207}\pm\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}}}}} \\ $$$$\mathrm{step}\:\mathrm{1} \\ $$$$\left({a}\pm{b}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{2207}\pm\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{a}=\frac{\mathrm{47}}{\mathrm{2}},\:{b}=\frac{\mathrm{21}}{\mathrm{2}} \\ $$$$\mathrm{step}\:\mathrm{2} \\ $$$$\left({a}\pm{b}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{47}\pm\mathrm{21}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{a}=\frac{\mathrm{7}}{\mathrm{2}},\:{b}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{step}\:\mathrm{3} \\ $$$$\left({a}\pm{b}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{7}\pm\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{a}=\frac{\mathrm{3}}{\mathrm{2}};\:{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{result}\:\mathrm{is} \\ $$$$\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ | ||
Commented by bhanukumarb2@gmail.com last updated on 01/Jun/19 | ||
$${thanks}\:{sir} \\ $$ | ||