Find the value of x in given equation: (√(3+x)+√(3−x))⁄(√(3+x)−√(3−x)) = 2

Correct — A. Let p = √(3+x) and q = √(3−x), so (p + q)⁄(p − q) = 2.

Cross-multiply: p + q = 2p − 2q
Collect terms: p = 3q

Square both sides: 3 + x = 9(3 − x)
Expand: 3 + x = 27 − 9x
Solve: 10x = 24, so x = 12⁄5 → option (a)

Why the others are wrong

• (b)
  7⁄5 — At x = 7⁄5, the radicands are 22⁄5 and 8⁄5, a ratio of 11⁄4, not the 9 that p = 3q needs. The fraction then comes to about 4.04, not 2.
• (c)
  6⁄5 — At x = 6⁄5, the radicands are 21⁄5 and 9⁄5, a ratio of 7⁄3 instead of 9. The fraction comes to about 4.79, far from 2.
• (d)
  8⁄5 — At x = 8⁄5, the radicands are 23⁄5 and 7⁄5, a ratio of 23⁄7 instead of 9. The fraction comes to about 3.46, not 2.

Componendo and dividendo: if a⁄b = c⁄d, then (a + b)⁄(a − b) = (c + d)⁄(c − d).

Applied to (p + q)⁄(p − q) = 2⁄1, it gives 2p⁄2q = (2 + 1)⁄(2 − 1), so p⁄q = 3 in one line, the same place cross-multiplying reaches.

Once p = 3q, squaring clears both roots together: the radicands must be in the ratio 9 : 1.

Key facts

• If a⁄b = c⁄d, then (a + b)⁄(a − b) = (c + d)⁄(c − d) (componendo and dividendo).
• If (p + q)⁄(p − q) = k, then p⁄q = (k + 1)⁄(k − 1).
• √(3 + x) and √(3 − x) are both real only when −3 ≤ x ≤ 3.

Common traps

• Squaring the fraction as it stands: the numerator becomes 6 + 2√(9 − x²), and the cross term turns a short solve into a long one.
• Stopping at p⁄q = 3 and writing (3 + x)⁄(3 − x) = 3: the radicands are in the ratio of the square, 9. The slip gives x = 3⁄2, which is not an option.

No directly related past PYQ was found.