Question

Find two positive even consecutive numbers, the sum of whose squares is 340.

Hint:

To solve problems involving consecutive numbers and their squares, use algebra to set up an equation and solve the resulting quadratic equation.

Answer 1

Let the two consecutive even numbers be \( x \) and \( x + 2 \).
Step 1: The sum of their squares is given by: \[ x^2 + (x + 2)^2 = 340. \]
Step 2: Expand the equation: \[ x^2 + (x^2 + 4x + 4) = 340. \]
Step 3: Simplify the equation: \[ 2x^2 + 4x + 4 = 340. \]
Step 4: Move all terms to one side: \[ 2x^2 + 4x + 4 - 340 = 0 \quad \Rightarrow \quad 2x^2 + 4x - 336 = 0. \]
Step 5: Divide the entire equation by 2 to simplify: \[ x^2 + 2x - 168 = 0. \]
Step 6: Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 1 \), \( b = 2 \), and \( c = -168 \). Substitute the values: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-168)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 672}}{2} = \frac{-2 \pm \sqrt{676}}{2}. \] Since \( \sqrt{676} = 26 \), we have: \[ x = \frac{-2 \pm 26}{2}. \]
Step 7: Thus, \( x = \frac{-2 + 26}{2} = \frac{24}{2} = 12 \) or \( x = \frac{-2 - 26}{2} = \frac{-28}{2} = -14 \). Since we are looking for positive even numbers, we take \( x = 12 \). Therefore, the two consecutive even numbers are \( 12 \) and \( 14 \).
Conclusion: The two consecutive even numbers are 12 and 14.