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I was doing a practice test for a math competition, and I came across a problem I am unable to solve, and I have tried to get help, with no luck. I need to be able to solve it without a calculator too. The problem is: "If $x^2+1/x^2=3$ and $x > 0$, what is the value of $x+1/x$. Express your answer in simplest radical form." I tried manipulating the first equation and I end up with $x^4-3x^2+1=0$, and there is no way I can figure out how to factor it. I would appreciate any help I can get with this problem!

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Guide:

Let $y=x^2$, solve for $y$ using the quadratic formula.

After you get your $y$, you can solve for the corresponding $x$.

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  • $\begingroup$ Can I get a more detailed explanation? I know the quadratic formula, but that doesn't really answer my question. $\endgroup$ – Spencer1O1 Jan 30 at 23:02
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    $\begingroup$ can you solve $y^2-3y+1=0$? $\endgroup$ – Siong Thye Goh Jan 31 at 0:53
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Notice that $x^2 \frac{1}{x^2} = 1$. We could add 2 to both sides of the equation to complete the square. $$ (x + \frac{1}{x})^2 = 5$$

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  • $\begingroup$ You mean $x^2 + \frac 1 {x^2}=3?$ $\endgroup$ – J. W. Tanner Jan 30 at 2:59
  • $\begingroup$ Thank you for pointing out the typo, I mean their product. It is corrected. @J.W.Tanner $\endgroup$ – Alvis Nordkovich Jan 30 at 3:04
  • $\begingroup$ Can I get a more detailed explanation please? $\endgroup$ – Spencer1O1 Jan 30 at 23:02
  • $\begingroup$ $$x^2 + \frac{1}{x^2} + 2 = 3 + 2 = 5$$ $$x^2 + x^2 \frac{1}{x^2} + \frac{1}{x^2} = 5$$ $$(x+\frac{1}{x})^2 = 5$$ $$x + \frac{1}{x} = \sqrt{5}$$ given $x > 0$ @Spencer1O1 $\endgroup$ – Alvis Nordkovich Jan 31 at 1:31
  • $\begingroup$ In my opinion, the OP would have benefited more from another hint than the full answer. $\endgroup$ – Toby Mak Aug 11 at 13:12

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