If x = 7 − 4√3, then find the value of (x² + 1⁄x²).

Correct — b. Rationalise 1⁄x first and the surds fall out.

1⁄x = 1⁄(7 − 4√3) × (7 + 4√3)⁄(7 + 4√3)
Denominator = 7² − (4√3)² = 49 − 48 = 1
So 1⁄x = 7 + 4√3

x + 1⁄x = (7 − 4√3) + (7 + 4√3) = 14

x² + 1⁄x² = (x + 1⁄x)² − 2
= 14² − 2
= 196 − 2 = 194 → option (b).

Why the others are wrong

• (a)
  188 — 188 is 196 − 8, the identity applied with 2ab where ab was taken as 4. But a × b here is x × 1⁄x = 1, so exactly 2 comes off.
• (c)
  186 — 186 is neither 14² nor 14² − 2. Once x + 1⁄x = 14 is fixed, the only numbers the identity can produce are 196 and 194.
• (d)
  196 — 196 is (x + 1⁄x)², the step before the last. The identity reads x² + 1⁄x² = (x + 1⁄x)² − 2, and that subtraction is not optional.

The question turns on one observation: 7 − 4√3 and 7 + 4√3 multiply to 1, because 7² − (4√3)² = 49 − 48.

That makes the reciprocal free — 1⁄x is simply the conjugate — and it makes x + 1⁄x a clean 14, the surd parts cancelling.

The identity then finishes it: x² + 1⁄x² = (x + 1⁄x)² − 2. The 2 is 2 × x × 1⁄x, and it is 2 precisely because that product is 1.

Key facts

• (7 − 4√3)(7 + 4√3) = 49 − 48 = 1, so the two numbers are reciprocals.
• With x = 7 − 4√3, x + 1⁄x = 14.
• x² + 1⁄x² = (x + 1⁄x)² − 2 = 196 − 2 = 194.
• 7 − 4√3 = (2 − √3)².

Common traps

• Stopping at (x + 1⁄x)² and answering 196
• Expanding (7 − 4√3)² and (7 + 4√3)² term by term, which works but costs a minute
• Writing (4√3)² as 12 instead of 48

No directly related past PYQ was found.