Question

The sum of the square of two consecutive positive even numbers (integers) is 340. Find the numbers.

Hint:

To solve problems involving consecutive numbers, assume the first number as \( x \) and express the next number in terms of \( x \).

Answer 1

Let the two consecutive even numbers be \( x \) and \( x + 2 \), where \( x \) is the first even number. According to the given condition, the sum of the squares of these numbers is 340: \[ x^2 + (x + 2)^2 = 340. \] Step 1: Expand the equation. Expanding \( (x + 2)^2 \): \[ x^2 + (x^2 + 4x + 4) = 340. \] Simplifying: \[ 2x^2 + 4x + 4 = 340. \] Step 2: Rearrange the equation. Move 340 to the left-hand side: \[ 2x^2 + 4x + 4 - 340 = 0 \quad \implies \quad 2x^2 + 4x - 336 = 0. \] Step 3: Divide the equation by 2. \[ x^2 + 2x - 168 = 0. \] Step 4: Solve the quadratic equation. Now, solve the quadratic equation \( x^2 + 2x - 168 = 0 \). Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 1 \), \( b = 2 \), and \( c = -168 \). Substituting the values: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-168)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 672}}{2} = \frac{-2 \pm \sqrt{676}}{2}. \] \[ x = \frac{-2 \pm 26}{2}. \] Step 5: Find the values of \( x \). We have two possible solutions for \( x \): \[ x = \frac{-2 + 26}{2} = \frac{24}{2} = 12 \quad \text{or} \quad x = \frac{-2 - 26}{2} = \frac{-28}{2} = -14. \] Since we are looking for positive even numbers, we choose \( x = 12 \). Thus, the two consecutive even numbers are \( 12 \) and \( 14 \).
Conclusion:
The two consecutive positive even numbers are \( 12 \) and \( 14 \).