If x^8 + 1/x^8 = 2207, what could be the value of x + 1/x ?

So one thing I haven't seen myself do too much particularly for the harder questions is recognize the quadratic identities. Is there a method to help you spot them? How did you quickly figure out that 47^2 = 2207?

Hello CEdward. Here we should know the expansion of \((a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2}+ 2\), which is derived from the expansion of \((a + b)^2 = a^2 + b^2 + 2ab\).

Therefore \(a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2\)

Now \(x^8 + \frac{1}{x^8} = (x^4 + \frac{1}{x^4})^2 - 2 = 2207\)

\((x^4 + \frac{1}{x^4})^2 = 2209\)

Now to find the square root of 2209, we know that \(50^2\) is 2500 and \(45^2\) (there is a short cut to find squares of numbers ending in 5) is 2025..So the square root should lie between between 45 and 50 and the square should end with 9. S0 \(47^2\) fits that.


\(x^4 + \frac{1}{x^4} = 47\)

\((x^2 + \frac{1}{x^2})^2 - 2 = 47\)

\((x^2 + \frac{1}{x^2})^2 = 49\)

\(x^2 + \frac{1}{x^2} = 7\)



Now, \((x + \frac{1}{x})^2 - 2 = 7\)

\((x + \frac{1}{x})^2 = 9\)

\(x + \frac{1}{x} = \pm 3\)


Hope this helps

Arun Kumar

CrackverbalGMAT Hi, could you explain why the square root considers both positive and negative value of 3 in this case? Thanks for the detailed solution posted above on 45-50 square root

For any value that is expressed as x^2 = n, x can be both a positive number and a negative number. So in the eventual answer of (x + 1/x)^2 = 9, both 3 and -3 should be considered. I'd imagine there would have been possible negative versions for ±47 and ±7, but were disregarded because they were not relevant to the answer we are seeking.