Simplify: (1⁄(7 + √2)) + (1⁄(7 − √2)) − (14⁄(49 − 2))

- (a)0
- (b)1
- (c)2
- (d)√2
Answer
Why
Correct — A. Add the first two fractions over one denominator.
Numerator: (7 − √2) + (7 + √2) = 14, the √2 terms cancel
Denominator: (7 + √2)(7 − √2) = 7² − (√2)² = 49 − 2 = 47
So the first two terms = 14⁄47
The third term is 14⁄(49 − 2) = 14⁄47
14⁄47 − 14⁄47 = 0 → option (a)
Why the others are wrong
- (b)1 — For 1, the first two fractions would have to total 1 + 14⁄47 = 61⁄47. They total 14⁄47, exactly the term being subtracted.
- (c)2 — For 2, the first pair would need to exceed 2. In decimals it is about 0.119 + 0.179 ≈ 0.298, which is 14⁄47 and nowhere near 2.
- (d)√2 — The √2 cancels in the numerator, (7 − √2) + (7 + √2) = 14, and the denominator 49 − 2 = 47 is rational, so no √2 survives.
Concept
7 + √2 and 7 − √2 are conjugates: the same two terms with the sign of the surd flipped. Their product is a difference of squares, (a + √b)(a − √b) = a² − b, so the surd disappears.
Adding their reciprocals gives 1⁄(a + √b) + 1⁄(a − √b) = 2a⁄(a² − b). With a = 7 and b = 2 that is 14⁄47.
The third term, 14⁄(49 − 2), is the sum of the first two written out, so the expression is built to cancel. Seeing that first saves the working.
Key facts
- (a + √b)(a − √b) = a² − b.
- 1⁄(a + √b) + 1⁄(a − √b) = 2a⁄(a² − b).
- (√2)² = 2, so (7 + √2)(7 − √2) = 49 − 2 = 47.
Study next
Common traps
- Writing (7 + √2)(7 − √2) as 49 − √2: the product subtracts the square of √2, which is 2.
- Adding the denominators: 1⁄(7 + √2) + 1⁄(7 − √2) is not 2⁄14, because fractions add over a common denominator.
Here the stem hands you a conjugate pair and prints their sum as the third term. A conjugate product also drives 21 Sep 2025, 16:00, Quant Q.24, where (5 + 2√6)(5 − 2√6) = 25 − 24 = 1 settles √(5 + 2√6) ÷ √(5 − 2√6).
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