If √x + 1⁄√x = 7, then the value of x + 1⁄x is equal to:

Correct — A. The stem is an image: √x + 1⁄√x = 7.

Square both sides, using (p + q)² = p² + 2pq + q² with p = √x and q = 1⁄√x:
(√x)² + 2 × √x × 1⁄√x + (1⁄√x)² = 7²
x + 2 + 1⁄x = 49

The middle term is 2 because √x × 1⁄√x = 1. That cancellation is the entire question.

Move the 2 across:
x + 1⁄x = 49 − 2 = 47 → option (a).

Why the others are wrong

• (b)
  51 — 51 is 49 + 2 — the cross term added instead of subtracted. Squaring puts +2 on the left of the equation, so it crosses over as a subtraction, not an addition.
• (c)
  45 — 45 is 49 − 4, the cross term taken as 2 × 2. The cross term is 2 × √x × 1⁄√x, and that product is 2 × 1 = 2.
• (d)
  49 — 49 is simply 7². It is the value of x + 2 + 1⁄x, so it overshoots the quantity asked for by exactly the cross term.

This is the square-of-a-sum identity in exam dress: (p + q)² = p² + q² + 2pq.

Choose p and q so that pq = 1 — here √x and 1⁄√x — and the cross term stops being an unknown and becomes the constant 2. That is the only reason this family of questions is solvable without finding x.

The ladder climbs. From √x + 1⁄√x you reach x + 1⁄x by squaring, then x² + 1⁄x² by squaring again, and x³ + 1⁄x³ by cubing.

Key facts

• (p + q)² = p² + q² + 2pq, and when pq = 1 the cross term is exactly 2.
• If √x + 1⁄√x = k then x + 1⁄x = k² − 2.
• If √x − 1⁄√x = k then x + 1⁄x = k² + 2.
• Squaring x + 1⁄x = m gives x² + 1⁄x² = m² − 2, the same step one rung higher.

Common traps

• Adding the 2 instead of subtracting it, which gives 51.
• Stopping at 49, the square of the expression you were given.
• Trying to solve for x first — it is not a whole number here, and you never need it.

No directly related past PYQ was found.