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Question-59752




Question Number 59752 by Khairun Nisa last updated on 14/May/19
Commented by Smail last updated on 14/May/19
((√(7+(√(48)))))^x +((√(7−(√(48)))))^x =14  ((√(7+(√(48)))))^x (((√(7+(√(48)))))^x +((√(7−(√(48)))))^x )=14((√(7+(√(48)))))^x   ((√(7+(√(48)))))^(2x) +((√((7+(√(48)))(7−(√(48))))))^x −14((√(7+(√(48)))))^x =0  (((√(7+(√(48)))))^x )^2 −14((√(7+(√(48)))))^x +1=0  (((√(7+(√(48)))))^x −7)^2 +1−49=0  ((√(7+(√(48)))))^x −7=+_− (√(48))  ((√(7+(√(48)))))^x =7+(√(48))  or =7−(√(48))  =((√(7+(√(48)))))^2   or =((1/( (√(7+(√(48)))))))^2 =((√(7+(√(48)))))^(−2)   x=2  or =−2
$$\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} +\left(\sqrt{\mathrm{7}−\sqrt{\mathrm{48}}}\right)^{{x}} =\mathrm{14} \\ $$$$\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} \left(\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} +\left(\sqrt{\mathrm{7}−\sqrt{\mathrm{48}}}\right)^{{x}} \right)=\mathrm{14}\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} \\ $$$$\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{\mathrm{2}{x}} +\left(\sqrt{\left(\mathrm{7}+\sqrt{\mathrm{48}}\right)\left(\mathrm{7}−\sqrt{\mathrm{48}}\right)}\right)^{{x}} −\mathrm{14}\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} =\mathrm{0} \\ $$$$\left(\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} \right)^{\mathrm{2}} −\mathrm{14}\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} +\mathrm{1}=\mathrm{0} \\ $$$$\left(\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} −\mathrm{7}\right)^{\mathrm{2}} +\mathrm{1}−\mathrm{49}=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} −\mathrm{7}=\underset{−} {+}\sqrt{\mathrm{48}} \\ $$$$\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{{x}} =\mathrm{7}+\sqrt{\mathrm{48}}\:\:{or}\:=\mathrm{7}−\sqrt{\mathrm{48}} \\ $$$$=\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{\mathrm{2}} \:\:{or}\:=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\right)^{−\mathrm{2}} \\ $$$${x}=\mathrm{2}\:\:{or}\:=−\mathrm{2} \\ $$
Commented by Khairun Nisa last updated on 14/May/19
Thanku Sir
$${Thanku}\:{Sir} \\ $$
Answered by tanmay last updated on 14/May/19
7+(√(48))   7+2(√(3×4))   =2^2 +((√3) )^2 +2×2×(√3)   =(2+(√3) )^2   (2+(√3) )^x +(2−(√3) )^x =14  (1/(2+(√3)))=2−(√3)   (2+(√3))^x =a  a+(1/a)=14  a^2 −14a+1=0  a=((14±(√(196−4)))/2)  ((14±8(√3))/2)=7±4(√3)   (2+(√3) )^x =(7+4(√3) )  (2+(√3) )^x =(2+(√3) )^2   x=2  (2+(√3) )^x =(7−4(√3) )  (2+(√3) )^x =(1/((7+4(√3) )))  (2+(√3) )^x =(2+(√3) )^(−2)   x=−2  so x=±2
$$\mathrm{7}+\sqrt{\mathrm{48}}\: \\ $$$$\mathrm{7}+\mathrm{2}\sqrt{\mathrm{3}×\mathrm{4}}\: \\ $$$$=\mathrm{2}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{2}×\sqrt{\mathrm{3}}\: \\ $$$$=\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{x}} +\left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right)^{{x}} =\mathrm{14} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\mathrm{2}−\sqrt{\mathrm{3}}\: \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} ={a} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{14} \\ $$$${a}^{\mathrm{2}} −\mathrm{14}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{14}\pm\sqrt{\mathrm{196}−\mathrm{4}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{14}\pm\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}\: \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{x}} =\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:\right) \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{x}} =\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} \\ $$$${x}=\mathrm{2} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{x}} =\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\:\right) \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{x}} =\frac{\mathrm{1}}{\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:\right)} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{x}} =\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{−\mathrm{2}} \\ $$$${x}=−\mathrm{2} \\ $$$${so}\:{x}=\pm\mathrm{2} \\ $$$$ \\ $$$$ \\ $$
Commented by Khairun Nisa last updated on 14/May/19
Thanku Sir
$${Thanku}\:{Sir} \\ $$
Commented by tanmay last updated on 14/May/19
most welcome
$${most}\:{welcome} \\ $$

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