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Question Number 7251 by Tawakalitu. last updated on 19/Aug/16

Find the value of x ((√(2 + (√3))))^x + ((√(2 − (√3))))^x = 4

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

Commented by sou1618 last updated on 19/Aug/16

(∗)∙∙∙(2+(√3))^(x/2) +(2−(√3))^(x/2) =4 (1/(2+(√3))) = (1/(2+(√3)))×((2−(√3))/(2−(√3))) = 2−(√3) (∗)⇒(2+(√3))^(x/2) +((1/(2+(√3))))^(x/2) =4 set y=(2+(√3))^(x/2) (y≠0) ⇒y+(1/y)=4 (y≠0) ⇒y^2 −4y+1=0 ⇒y=2±(√(4−1)) =2±(√3) (2+(√3))^(x/2) =2+(√3),2−(√3) (2+(√3))^(x/2) =2+(√3),(1/(2+(√3))) (x/2)=1,−1 x=±2

$$\left(\ast\right)\centerdot\centerdot\centerdot\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}/\mathrm{2}} +\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}/\mathrm{2}} =\mathrm{4} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}×\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\left(\ast\right)\Rightarrow\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}/\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}/\mathrm{2}} =\mathrm{4} \\ $$$${set}\:{y}=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}/\mathrm{2}} \:\:\:\left({y}\neq\mathrm{0}\right) \\ $$$$\Rightarrow{y}+\frac{\mathrm{1}}{{y}}=\mathrm{4}\:\:\:\:\left({y}\neq\mathrm{0}\right) \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}/\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}},\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}/\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\frac{{x}}{\mathrm{2}}=\mathrm{1},−\mathrm{1} \\ $$$${x}=\pm\mathrm{2} \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 19/Aug/16

Nice!

$$\mathcal{N}{ice}! \\ $$

Commented by Tawakalitu. last updated on 19/Aug/16

Wow, thanks so much sir, God bless you.

$${Wow},\:{thanks}\:{so}\:{much}\:{sir},\:{God}\:{bless}\:{you}. \\ $$

Commented by peter james last updated on 19/Aug/16

very nice...

$${very}\:{nice}... \\ $$

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