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GATE CSE 2021 Set 2 Q4: Solve a difference of two squares equal to x + 2

Question

If (x−12)2−(x−32)2=x+2\left(x - \dfrac{1}{2}\right)^2 - \left(x - \dfrac{3}{2}\right)^2 = x + 2, then the value of xx is:

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Answer: B (official GATE 2021 final answer key)

Concept used: Difference of squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).

First step: Here a−b=1a - b = 1 and a+b=2x−2a + b = 2x - 2, so the left side is 2x−22x - 2; set it equal to x+2x + 2.

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Source: official GATE 2021 Set 2 CS question paper and final answer key, organised by IIT Bombay. Question paper · Answer key

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