Algebraic identity with radicals (minimal repair for clarity): Let x = √a + 1 and y = √a − 1 with a > 0. Evaluate the expression (x^4 + y^4 − 2x^2 y^2) / [ (√a)(√a) ].

Difficulty: Medium

Correct Answer: 16

Explanation:


Introduction / Context:
Expressions like x^4 + y^4 − 2x^2 y^2 simplify dramatically using the identity x^4 + y^4 − 2x^2 y^2 = (x^2 − y^2)^2. With x and y written in terms of √a, the task reduces to computing a simple difference and then dividing by (√a)(√a) = a.



Given Data / Assumptions:

  • x = √a + 1, y = √a − 1, with a > 0.
  • Compute (x^4 + y^4 − 2x^2 y^2) / a.


Concept / Approach:
First compute x^2 − y^2, because the target expression equals (x^2 − y^2)^2 / a. Squaring binomials with radicals requires care but is routine. Finally, divide by a to obtain a constant.



Step-by-Step Solution:

x^2 = (√a + 1)^2 = a + 2√a + 1y^2 = (√a − 1)^2 = a − 2√a + 1x^2 − y^2 = (a + 2√a + 1) − (a − 2√a + 1) = 4√aTherefore, x^4 + y^4 − 2x^2 y^2 = (x^2 − y^2)^2 = (4√a)^2 = 16aDivide by a: (16a)/a = 16


Verification / Alternative check:
Recognize symmetry: replacing 1 by −1 in y changes only the middle term of (√a ± 1)^2, making the difference double that middle term; squaring then removes the radical.



Why Other Options Are Wrong:
5, 10, and 20 stem from arithmetic slips like squaring 4√a incorrectly (e.g., writing 4a or 8a) or forgetting to divide by a.



Common Pitfalls:
Miscomputing (√a ± 1)^2 or neglecting that (√a)(√a) = a leads to wrong constants.



Final Answer:
16

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion