2-3-x-2-2-3-x-2-4-

Question Number 103914 by bobhans last updated on 18/Jul/20

$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:+\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:=\:\mathrm{4}\: \\ $$

Commented by som(math1967) last updated on 18/Jul/20

let (2+(√3))^x^2 =a ∴(2−(√3))^x^2 =(1/a) ⇒a^2 −4a+1=0 ⇒a=(2±(√3)) (2+(√3))^x^2 =(2+(√3)) ⇒x=±1 (2+(√3))^x^2 =(2+(√3))^(−1) x=i

$$\mathrm{let}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}^{\mathrm{2}} } =\mathrm{a} \\ $$$$\therefore\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}^{\mathrm{2}} } =\frac{\mathrm{1}}{\mathrm{a}} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{4a}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{a}=\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}^{\mathrm{2}} } =\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{x}=\pm\mathrm{1} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}^{\mathrm{2}} } =\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{−\mathrm{1}} \\ $$$$\mathrm{x}=\mathrm{i} \\ $$

Commented by bemath last updated on 18/Jul/20

$${thank}\:{you}\:{both} \\ $$

Answered by OlafThorendsen last updated on 18/Jul/20

X = (2+(√3))^x^2 (2−(√3))^x^2 = ((1/(2+(√3))))^x^2 = (1/((2+(√3))^x^2 )) = (1/X) then X+(1/X) = 4 X^2 −4X+1 = 0 Δ = (−4)^2 −4(1)(1) = 12 X = ((−(−4)±(√(12)))/(2(1))) = 2±(√3) 1) X = 2+(√3) x^2 = 1, x=±1 2) X = 2−(√3) 2−(√3) = (2+(√3))^x^2 (1/(2+(√3))) = (2+(√3))^x^2 (2+(√3))^(x^2 +1) = 1 x^2 +1 = 0 x = ±i S = {±1;±i}

$${X}\:=\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:=\:\left(\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}^{\mathrm{2}} } =\:\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } }\:=\:\frac{\mathrm{1}}{{X}} \\ $$$$\mathrm{then}\:{X}+\frac{\mathrm{1}}{{X}}\:=\:\mathrm{4} \\ $$$${X}^{\mathrm{2}} −\mathrm{4}{X}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Delta\:=\:\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)\:=\:\mathrm{12} \\ $$$${X}\:=\:\frac{−\left(−\mathrm{4}\right)\pm\sqrt{\mathrm{12}}}{\mathrm{2}\left(\mathrm{1}\right)}\:=\:\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\left.\mathrm{1}\right)\:{X}\:=\:\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{1},\:{x}=\pm\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{X}\:=\:\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\mathrm{2}−\sqrt{\mathrm{3}}\:=\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\:=\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} +\mathrm{1}} \:=\:\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}\:=\:\pm{i} \\ $$$$\mathscr{S}\:=\:\left\{\pm\mathrm{1};\pm{i}\right\} \\ $$

Answered by Rauny last updated on 18/Jul/20

(2+(√3))^x^2 + (2−(√3))^x^2 = 4 t:=2+(√3) (1/(2+(√3)))=((2−(√3))/((2+(√3))(2−(√3))))=2−(√3) ∴2−(√3)=(1/t) t^x^2 +t^(−x^2 ) −4=0 u:=t^x^2 (∀u>0) u−4+(1/u)=0 u^2 −4u+1=0 u=2±(√3) t^x^2 =2±(√3) x^2 =log_(2+(√3)) (2±(√3)) if u=2+(√3), x^2 =1 x=±1 if u=2−(√3), x^2 =log_(2+(√3)) (1/(2+(√3)))=−1 (∵2−(√3)=(1/(2+(√3)))) x=±i ∴x=±1 or ±i ■

$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:+\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:=\:\mathrm{4}\: \\ $$$${t}:=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\therefore\mathrm{2}−\sqrt{\mathrm{3}}=\frac{\mathrm{1}}{{t}} \\ $$$${t}^{{x}^{\mathrm{2}} } +{t}^{−{x}^{\mathrm{2}} } −\mathrm{4}=\mathrm{0} \\ $$$${u}:={t}^{{x}^{\mathrm{2}} } \:\left(\forall{u}>\mathrm{0}\right) \\ $$$${u}−\mathrm{4}+\frac{\mathrm{1}}{{u}}=\mathrm{0} \\ $$$${u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}=\mathrm{0} \\ $$$${u}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${t}^{{x}^{\mathrm{2}} } =\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} =\mathrm{log}_{\mathrm{2}+\sqrt{\mathrm{3}}} \:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{if}\:{u}=\mathrm{2}+\sqrt{\mathrm{3}}, \\ $$$${x}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}=\pm\mathrm{1} \\ $$$$\mathrm{if}\:{u}=\mathrm{2}−\sqrt{\mathrm{3}}, \\ $$$${x}^{\mathrm{2}} =\mathrm{log}_{\mathrm{2}+\sqrt{\mathrm{3}}} \:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}=−\mathrm{1}\:\left(\because\mathrm{2}−\sqrt{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right) \\ $$$${x}=\pm{i} \\ $$$$\therefore{x}=\pm\mathrm{1}\:\mathrm{or}\:\pm{i}\:\blacksquare \\ $$

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