Question

Sum of the squares of two consecutive even positive integers is 340. The numbers are

• A. 10, 12
• B. 12, 16
• C. 14, 16
• D. 12, 14

Answer 1

The Correct Option is D

Let the two consecutive even integers be \( x \) and \( x + 2 \). Then, their squares will be \( x^2 \) and \( (x+2)^2 \). According to the question: \[ x^2 + (x+2)^2 = 340 \] \[ x^2 + x^2 + 4x + 4 = 340 \] \[ 2x^2 + 4x + 4 = 340 \] \[ 2x^2 + 4x - 336 = 0 \] Divide the whole equation by 2: \[ x^2 + 2x - 168 = 0 \] Solve the quadratic: \[ x = \frac{-2 \pm \sqrt{(2)^2 + 4 \cdot 168}}{2} = \frac{-2 \pm \sqrt{4 + 672}}{2} = \frac{-2 \pm \sqrt{676}}{2} = \frac{-2 \pm 26}{2} \] So, \( x = 12 \) or \( x = -14 \) Only the positive even integer is valid: \( x = 12 \)

Thus, the numbers are 12 and 14