If √(1 + 27/169) = (1 + x/13), then the value of x is

Concept: The radical sign √ denotes the principal square root, which is by definition non-negative: for a ≥ 0, √a is the unique number that is both non-negative and squares to a. An equation of the form √A = B therefore carries two conditions at once — B must satisfy B² = A, and B must itself be non-negative. Squaring both sides drops the second condition, which is how extra roots that the original radical equation rejects can appear.

Application to this equation:

1. Combine the quantity under the radical: 1 + 27/169 = 169/169 + 27/169 = 196/169.

2. Take the principal square root: 196 = 14² and 169 = 13², so √(196/169) = 14/13, a positive number.

3. The equation becomes 1 + x/13 = 14/13.

4. Write the left side over a common denominator: (13 + x)/13 = 14/13, so 13 + x = 14.

5. Therefore x = 1.

Cross-check every listed value of x:

• x = 1 gives 1 + x/13 = 14/13. This is non-negative and its square is 196/169, so it satisfies the printed equation.

• x = −27 gives 1 + x/13 = −14/13. Its square is also 196/169, so it satisfies the squared equation (1 + x/13)² = 196/169, but −14/13 is negative while the principal square root of 196/169 is +14/13, so it does not satisfy the printed equation.

• x = −14/13 gives 1 + x/13 = 1 − 14/169 = 155/169, which is neither 14/13 nor −14/13.

Exactly one listed value, x = 1, satisfies the given equation, so the value of x is 1. The value −27 is the classic trap: it is a root of the squared equation only, and it is discarded because a principal square root can never be negative.

Note for BPSC aspirants: the commission’s official key for this paper marked the “more than one of the above” choice, which is obtainable only by solving the squared equation and keeping its negative branch. As the printed equation uses the radical sign, the mathematically defensible value is x = 1, and standard solved-paper sources treat the official key for this item as an error.