Given, x + 1⁄x = 5, then determine the value of x² + 1⁄x².

Correct — D. Square the given equation, then remove the middle term.

(x + 1⁄x)² = x² + 2·x·(1⁄x) + 1⁄x² = x² + 2 + 1⁄x²
So x² + 2 + 1⁄x² = 5² = 25
Subtract 2: x² + 1⁄x² = 25 − 2 = 23 → option (d)

Why the others are wrong

• (a)
  25 — 25 is just 5², the square of x + 1⁄x. That square still holds the middle term 2·x·(1⁄x) = 2, which has to come off.
• (b)
  24 — 24 takes off 1 instead of 2. The middle term is 2 × x × 1⁄x, and x × 1⁄x = 1, so it is exactly 2.
• (c)
  26 — 26 adds 1 to 25 instead of subtracting 2. Squaring x + 1⁄x brings in an extra 2, which must be subtracted, not added to.

When x + 1⁄x is known, the square identity gives the next power: (x + 1⁄x)² = x² + 1⁄x² + 2. The 2 is x × 1⁄x = 1, counted twice.

So x² + 1⁄x² = (x + 1⁄x)² − 2. Repeat the step on x² + 1⁄x² to get x⁴ + 1⁄x⁴: here 23² − 2 = 527.

Key facts

• x² + 1⁄x² = (x + 1⁄x)² − 2.
• x² + 1⁄x² = (x − 1⁄x)² + 2.
• x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x).

Common traps

• Stopping at 5² = 25 and forgetting the middle term
• Subtracting 2 when the given sum is x − 1⁄x, where the 2 is added instead

No directly related past PYQ was found.