Question:

Two consecutive even positive integers, sum of the squares of which is 1060 are

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For sum of squares problems, form a quadratic equation and solve for the integers.
Updated On: Aug 18, 2025
  • 12 and 14
  • 22 and 24
  • 20 and 22
  • 16 and 18
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The Correct Option is B

Solution and Explanation

Let the two consecutive even integers be \( x \) and \( x + 2 \). The sum of the squares of the integers is given as: \[ x^2 + (x + 2)^2 = 1060 \] Expanding the equation: \[ x^2 + x^2 + 4x + 4 = 1060 \] \[ 2x^2 + 4x + 4 = 1060 \] \[ 2x^2 + 4x - 1056 = 0 \] Divide the whole equation by 2: \[ x^2 + 2x - 528 = 0 \] Using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-528)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 2112}}{2} = \frac{-2 \pm \sqrt{2116}}{2} = \frac{-2 \pm 46}{2} \] Thus, \( x = 22 \), and the consecutive integers are 22 and 24. - Option (A) 12 and 14: Incorrect. The square sum does not match for these values.
- Option (C) 20 and 22: Incorrect. This does not satisfy the equation.
- Option (D) 16 and 18: Incorrect. The sum of squares is too low.
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