Square both side, one square root remains from the cross-multiplication on the left side. Put everything else on the other side, square again, and you now have a 4th degree polynomial to solve (thankfully the 1/x and 1/x² cancel). I even worked it out and here it is: x^4 - 2 x^3 - x² +2x +1 = 0 Now, this does factor as (x² - x -1)² = 0, and we're back on track for the solution, but that factorization is not easy to see unless you already know where you are going :/ The palindrome in the coefficient is an hint, but not much of one.
Are you kidding man… There are many mistakes in your solution. When you are multiplying both sides by (a - b) you should check whether it is zero. Also when you have found x you should check it by putting it into the original equation (cause there could be some issues with b < 0 etc)
Firstly, no need to focus on checking whether (a-b) is equal to 0, because if it is, then a = b, that is x - 1/x = 1 - 1/x. This in turn means that x = 1, but if we put this value into the original equation, we will get 0 = 1, which is obviously wrong Secondly, it is too obvious for me to put the value of x into the original equation there (1.618 > 1, so 1/618 < 1, and that means b is positive)
