I need help solving a competition math problem.

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!

Guide:

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

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

• Can I get a more detailed explanation? I know the quadratic formula, but that doesn't really answer my question. – Spencer1O1 Jan 30 at 23:02
• can you solve $y^2-3y+1=0$? – Siong Thye Goh Jan 31 at 0:53

Notice that $$x^2 \frac{1}{x^2} = 1$$. We could add 2 to both sides of the equation to complete the square. $$(x + \frac{1}{x})^2 = 5$$

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