x^2 + (1/x^2 ) = 51 find x


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Question Number 83672 by jagoll last updated on 05/Mar/20

$x^{2} + \frac{1}{x^{2}} = 51$ $f i n d x$
	Answered by john santu last updated on 05/Mar/20
	$x^2 + (1/x^2 ) = 51 ⇔ (x+(1/x))^2 −2 = x^2 +(1/x^2 ) (x+(1/x))^2 = 53 ⇒x+(1/x)= ± (√(53)) case (1) ⇒x+(1/x) = (√(53)) x^2 −x(√(53)) +1 = 0 x = (((√(53)) ± 7)/2) = { ((x= (((√(53)) +7)/2))),((x = (((√(53)) −7)/2))) :} case (2) ⇒x +(1/x) = −(√(53)) x^2 +x(√(53)) +1 =0 x = ((−(√(53)) ± 7)/2) = { ((x= ((7−(√(53)))/2))),((x = ((−7−(√(53)))/2))) :}$
	$x^{2} + \frac{1}{x^{2}} = 51 \Leftrightarrow {(x + \frac{1}{x})}^{2} - 2 = x^{2} + \frac{1}{x^{2}}$ ${(x + \frac{1}{x})}^{2} = 53 \Rightarrow x + \frac{1}{x} = \pm \sqrt{53}$ $case (1) \Rightarrow x + \frac{1}{x} = \sqrt{53}$ $x^{2} - x \sqrt{53} + 1 = 0$ $x = \frac{\sqrt{53} \pm 7}{2} = {\begin{cases} x = \frac{\sqrt{53} + 7}{2} \\ x = \frac{\sqrt{53} - 7}{2} \end{cases}$ $case (2) \Rightarrow x + \frac{1}{x} = - \sqrt{53}$ $x^{2} + x \sqrt{53} + 1 = 0$ $x = \frac{- \sqrt{53} \pm 7}{2} = {\begin{cases} x = \frac{7 - \sqrt{53}}{2} \\ x = \frac{- 7 - \sqrt{53}}{2} \end{cases}$
	Answered by mr W last updated on 05/Mar/20
	$(x + (1/x))^2 = 53 ⇒x+(1/x)=±(√(53)) (x − (1/x))^2 = 49 ⇒x−(1/x)=±7 ⇒x=(1/2)(±(√(53))±7)= { ((((√(53))+7)/2)),((((√(53))−7)/2)),(((−(√(53))+7)/2)),(((−(√(53))−7)/2)) :}$
	${(x + \frac{1}{x})}^{2} = 53 \Rightarrow x + \frac{1}{x} = \pm \sqrt{53}$ ${(x - \frac{1}{x})}^{2} = 49 \Rightarrow x - \frac{1}{x} = \pm 7$ $\Rightarrow x = \frac{1}{2} (\pm \sqrt{53} \pm 7) = {\begin{cases} \frac{\sqrt{53} + 7}{2} \\ \frac{\sqrt{53} - 7}{2} \\ \frac{- \sqrt{53} + 7}{2} \\ \frac{- \sqrt{53} - 7}{2} \end{cases}$
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