Question Number 9141 by tawakalitu last updated on 20/Nov/16
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{if} \\ $$$$\left(\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} \:+\:\left(\sqrt{\mathrm{2}\:−\:\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} \:=\:\mathrm{4} \\ $$
Commented by tawakalitu last updated on 20/Nov/16
$$\mathrm{please}\:\mathrm{help}. \\ $$
Commented by RasheedSoomro last updated on 21/Nov/16
$$\mathrm{x}=\mathrm{2} \\ $$
Commented by tawakalitu last updated on 21/Nov/16
$$\mathrm{please}\:\mathrm{working} \\ $$
Commented by sou1618 last updated on 21/Nov/16
$$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}} \\ $$$${so} \\ $$$$\Leftrightarrow\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} +\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{−{x}} =\mathrm{4} \\ $$$${t}=\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} \\ $$$$\Rightarrow{t}+\frac{\mathrm{1}}{{t}}=\mathrm{4} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{2}\pm\sqrt{\mathrm{3}}=\mathrm{2}+\sqrt{\mathrm{3}}\:,\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${x}=\mathrm{2},−\mathrm{2} \\ $$$$ \\ $$
Commented by tawakalitu last updated on 21/Nov/16
$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$