For a positive real number a, if a = 1/(a − 5), what is the value of a + 1/a?

Introduction / Context:
This algebraic question involves solving a rational equation in one variable and then evaluating a symmetric expression a + 1/a. Such questions are common in aptitude tests because they explore quadratic equation solving skills and manipulation of reciprocal expressions without explicitly requiring full solution of all roots when additional conditions are given.

Given Data / Assumptions:

• a is a positive real number.
• a = 1/(a − 5).
• We need to find a + 1/a.

Concept / Approach:
First we clear the denominator to obtain a quadratic equation in a. Then we solve the quadratic and use the condition that a is positive to choose the correct root. Finally, we compute a + 1/a and simplify it to a square root form. Notice that we do not need exact decimal values; the exact radical form is enough.

Step-by-Step Solution:
Start from a = 1/(a − 5). Multiply both sides by (a − 5) to clear the denominator: a(a − 5) = 1. Expand: a^2 − 5a = 1. Rearrange to standard quadratic form: a^2 − 5a − 1 = 0. Use the quadratic formula: a = [5 ± √(25 + 4)] / 2 = [5 ± √29] / 2. Compute approximate values to decide the sign: √29 is about 5.385, so (5 + √29)/2 is positive and greater than 5, while (5 − √29)/2 is negative. Given that a is positive, we take a = (5 + √29)/2. Now compute a + 1/a. Instead of direct substitution, use the original quadratic: a^2 − 5a − 1 = 0 implies a^2 = 5a + 1. Then a + 1/a = (a^2 + 1)/a = (5a + 1 + 1)/a = (5a + 2)/a = 5 + 2/a. Substitute a = (5 + √29)/2 and compute 2/a = 4/(5 + √29). Rationalise: 4/(5 + √29) = 4(5 − √29)/(25 − 29) = 4(5 − √29)/(−4) = −5 + √29. Therefore a + 1/a = 5 + (−5 + √29) = √29.
Verification / Alternative check:
You can directly substitute the approximate numeric value of a into a + 1/a. With a ≈ 5.1926, compute a + 1/a ≈ 5.1926 + 0.1926 ≈ 5.3852, which is very close to √29, confirming the result.

Why Other Options Are Wrong:
The option −√29 corresponds to the negative root of the quadratic, which is discarded by the condition that a is positive. The option √(−29) is not real. √27 and 5 have no algebraic support from the quadratic equation and arise only from guesswork or arithmetic errors.

Common Pitfalls:
Students sometimes forget to use the positivity condition and keep both roots, leading to ambiguity. Another frequent mistake is incorrectly rationalising 4/(5 + √29). Carefully applying the quadratic formula and checking signs at each stage avoid these issues.

Final Answer:
The required value of a + 1/a is √29.