If x⁴ + 1/x⁴ = 527, then the value of x + 1/x is

Concept

For any nonzero x, the symmetric powers are linked by repeated squaring of the basic sum s = x + 1/x. Squaring a sum adds the cross term, so (a + b)2 = a2 + 2ab + b2. With b = 1/a the middle term is the constant 2, which lets each higher power be reached from the one below.

Application

1. Let s = x + 1/x. Squaring: s2 = x2 + 2 + 1/x2, so x2 + 1/x2 = s2 - 2.

2. Square again: (x2 + 1/x2)2 = x4 + 2 + 1/x4, so x4 + 1/x4 = (x2 + 1/x2)2 - 2.

3. Substitute: 527 = (x2 + 1/x2)2 - 2, so (x2 + 1/x2)2 = 529 and x2 + 1/x2 = ±23.

4. For real x, x2 + 1/x2 ≥ 2 (a square plus its reciprocal is always at least 2 by AM-GM), so the -23 branch is impossible and x2 + 1/x2 = 23.

5. Back to step 1: s2 - 2 = 23, so s2 = 25 and s = x + 1/x = ±5.

Choosing the value

s2 = 25 admits both +5 and -5 (for real x, +5 corresponds to x > 0 and -5 to x < 0). Only positive values are offered here, so the answer is 5, matching the standard convention of taking the positive root for this problem type.

Cross-check

Forward from s = 5: x2 + 1/x2 = 52 - 2 = 23, and x4 + 1/x4 = 232 - 2 = 527, matching the given value (s = -5 gives the same fourth power, since only even powers appear).