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Question Number 170485 by Mastermind last updated on 24/May/22

Answered by Rasheed.Sindhi last updated on 25/May/22

((√(2+(√3))) )^x +((√(2−(√3))) )^x =2 ; x=? (1/(((√(2+(√3))) )^x ))=(1/(((√(2+(√3))) )^x ))∙((((√(2−(√3))) )^x )/(((√(2−(√3))) )^x )) =((((√(2−(√3))) )^x )/(((√(4−3)))^x ))=((√(2−(√3))) )^x ((√(2+(√3))) )^x +(1/(((√(2+(√3))) )^x ))=2 a+(1/a)=2 a^2 −2a+1=0 (a−1)^2 =0 a=1 ((√(2+(√3))) )^x =1=((√(2+(√3))) )^0 x=0

$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} +\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} =\mathrm{2}\:;\:{x}=? \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} }=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} }\centerdot\frac{\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} }{\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} } \\ $$$$\:\:=\frac{\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} }{\left(\sqrt{\mathrm{4}−\mathrm{3}}\right)^{{x}} }=\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} +\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} }=\mathrm{2} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{2} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\mathrm{1} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} =\mathrm{1}=\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{\mathrm{0}} \\ $$$${x}=\mathrm{0} \\ $$

Answered by Rasheed.Sindhi last updated on 25/May/22

((√(2+(√3))))^x +((√(2−(√3))))^x =2 ; x=? Multiplying by ((√(2+(√3))))^x to both sides: ((√(2+(√3))))^(2x) +((√(2+(√3)))∙(√(2−(√3))))^x =2((√(2+(√3))))^x ((√(2+(√3))))^(2x) +((√(4−3)))^x =2((√(2+(√3))))^x Let ((√(2+(√3))))^x =a a^2 −2a+1=0 (a−1)^2 =0 a=1 ((√(2+(√3))))^x =1=((√(2+(√3))))^0 x=0

$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} +\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{{x}} =\mathrm{2}\:;\:{x}=? \\ $$$${Multiplying}\:{by}\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} {to}\:{both}\:{sides}: \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{\mathrm{2}{x}} +\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\centerdot\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{{x}} =\mathrm{2}\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{\mathrm{2}{x}} +\left(\sqrt{\mathrm{4}−\mathrm{3}}\right)^{{x}} =\mathrm{2}\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} \\ $$$${Let}\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} ={a} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\mathrm{1} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} =\mathrm{1}=\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{\mathrm{0}} \\ $$$${x}=\mathrm{0} \\ $$

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