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Question Number 83672 by jagoll last updated on 05/Mar/20
x^2 + (1/x^2 ) = 51 find x
$${x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\mathrm{51}\: \\ $$$${find}\:{x}\: \\ $$
Answered by john santu last updated on 05/Mar/20
x^2 + (1/x^2 ) = 51 ⇔ (x+(1/x))^2 −2 = x^2 +(1/x^2 ) (x+(1/x))^2 = 53 ⇒x+(1/x)= ± (√(53)) case (1) ⇒x+(1/x) = (√(53)) x^2 −x(√(53)) +1 = 0 x = (((√(53)) ± 7)/2) = { ((x= (((√(53)) +7)/2))),((x = (((√(53)) −7)/2))) :} case (2) ⇒x +(1/x) = −(√(53)) x^2 +x(√(53)) +1 =0 x = ((−(√(53)) ± 7)/2) = { ((x= ((7−(√(53)))/2))),((x = ((−7−(√(53)))/2))) :}
$${x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\mathrm{51}\:\Leftrightarrow\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\:=\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\:\mathrm{53}\:\Rightarrow{x}+\frac{\mathrm{1}}{{x}}=\:\pm\:\sqrt{\mathrm{53}} \\ $$$$\mathrm{case}\:\left(\mathrm{1}\right)\:\Rightarrow{x}+\frac{\mathrm{1}}{{x}}\:=\:\sqrt{\mathrm{53}} \\ $$$${x}^{\mathrm{2}} −{x}\sqrt{\mathrm{53}}\:+\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{\sqrt{\mathrm{53}}\:\pm\:\mathrm{7}}{\mathrm{2}}\:=\:\begin{cases}{{x}=\:\frac{\sqrt{\mathrm{53}}\:+\mathrm{7}}{\mathrm{2}}}\\{{x}\:=\:\frac{\sqrt{\mathrm{53}}\:−\mathrm{7}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{case}\:\left(\mathrm{2}\right)\:\Rightarrow{x}\:+\frac{\mathrm{1}}{{x}}\:=\:−\sqrt{\mathrm{53}} \\ $$$${x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{53}}\:+\mathrm{1}\:=\mathrm{0} \\ $$$${x}\:=\:\frac{−\sqrt{\mathrm{53}}\:\pm\:\mathrm{7}}{\mathrm{2}}\:=\:\begin{cases}{{x}=\:\frac{\mathrm{7}−\sqrt{\mathrm{53}}}{\mathrm{2}}}\\{{x}\:=\:\frac{−\mathrm{7}−\sqrt{\mathrm{53}}}{\mathrm{2}}}\end{cases} \\ $$
Answered by mr W last updated on 05/Mar/20
(x + (1/x))^2 = 53 ⇒x+(1/x)=±(√(53)) (x − (1/x))^2 = 49 ⇒x−(1/x)=±7 ⇒x=(1/2)(±(√(53))±7)= { ((((√(53))+7)/2)),((((√(53))−7)/2)),(((−(√(53))+7)/2)),(((−(√(53))−7)/2)) :}
$$\left({x}\:+\:\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:=\:\mathrm{53}\:\Rightarrow{x}+\frac{\mathrm{1}}{{x}}=\pm\sqrt{\mathrm{53}} \\ $$$$\left({x}\:−\:\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:=\:\mathrm{49}\:\Rightarrow{x}−\frac{\mathrm{1}}{{x}}=\pm\mathrm{7} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pm\sqrt{\mathrm{53}}\pm\mathrm{7}\right)=\begin{cases}{\frac{\sqrt{\mathrm{53}}+\mathrm{7}}{\mathrm{2}}}\\{\frac{\sqrt{\mathrm{53}}−\mathrm{7}}{\mathrm{2}}}\\{\frac{−\sqrt{\mathrm{53}}+\mathrm{7}}{\mathrm{2}}}\\{\frac{−\sqrt{\mathrm{53}}−\mathrm{7}}{\mathrm{2}}}\end{cases} \\ $$

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