The difference of the squares of two positive consecutive even integes is 68 . Find the integers. Use the fat that, if \( x \) represents an even integer, ther \( x+2 \) represents the next consecutiv integer.

Let's set up the problem using the information given. Let the first even integer be \( x \). Then, the next consecutive even integer is \( x + 2 \). The difference of their squares can be expressed as: \[ (x + 2)^2 - x^2 = 68 \] Expanding both squares, we have: \[ (x^2 + 4x + 4) - x^2 = 68 \] This simplifies to: \[ 4x + 4 = 68 \] Subtracting 4 from both sides: \[ 4x = 64 \] Dividing by 4 gives us: \[ x = 16 \] So, the two positive consecutive even integers are \( 16 \) and \( 18 \). Check: \[ (18^2 - 16^2) = (324 - 256) = 68. \] This confirms that our integers are indeed \( 16 \) and \( 18 \). Now, if you're curious about how to approach these types of problems effectively, remember that you can always set up an equation based on the relationships provided. Using the identity of the difference of squares \( a^2 - b^2 = (a - b)(a + b) \) can make it much simpler! And for a bit of historical context, algebraic equations such as this one have roots tracing back to ancient civilizations. The Babylonians were solving quadratic-like problems around 2000 BC, laying the groundwork for the algebra we use today. It's fascinating how these mathematical ideas have evolved!