The product of two consecutive even integers is 8 less than five times their sum. Find the two integers

Let the two consecutive even integers be x and x+2.

Their product is given by:

x(x+2)

The sum of the integers is:

x + (x+2) = 2x + 2

Five times their sum is:

5(2x + 2) = 10x + 10

According to the problem, the product is 8 less than five times their sum:

x(x+2) = 10x + 10 - 8

Simplifying, we have:

x(x+2) = 10x + 2

This expands to:

x^2 + 2x = 10x + 2

Rearranging, we get:

x^2 + 2x - 10x - 2 = 0

Simplifying, we have:

x^2 - 8x - 2 = 0

Solve this quadratic equation using the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a = 1, b = -8, and c = -2:

x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-2)}}{2(1)}

x = \frac{8 \pm \sqrt{64 + 8}}{2}

x = \frac{8 \pm \sqrt{72}}{2}

x = \frac{8 \pm \sqrt{36 \times 2}}{2}

x = \frac{8 \pm 6\sqrt{2}}{2}

Continuing to simplify gives us:

x = 4 \pm 3\sqrt{2}

Since the solutions involve square roots, there are no integer solutions to the problem.