Find two consecutive even intergers such that the square of the smaller is 10 more than the larger.

first number --- x

next consecutive even number --- x+2

x^2= x+2 + 10
x^2 - x - 12 = 0
(x-4)(x+3) = 0
x = 4 or x = -3 , but x has to be even

x = 4
numbers are 4 and 6

check: square of smaller is 16
10 more than the larger is 10+6 = 16
Yeah!

To find two consecutive even integers, we need to define two variables. Let's call the smaller even integer "x", and the larger even integer "x + 2" since they are consecutive.

According to the problem, the square of the smaller integer is 10 more than the larger integer. Mathematically, this can be expressed as:

x^2 = (x + 2) + 10

To solve this equation, we need to simplify and solve for "x".

Expanding the equation:

x^2 = x + 12

Rearranging the equation to have zero on one side:

x^2 - x - 12 = 0

Next, we can factor this quadratic equation:

(x - 4)(x + 3) = 0

This gives us two possible values for "x": x = 4 and x = -3. However, we are looking for even integers, so we disregard the negative value.

Therefore, x = 4 is the smaller even integer.

To find the larger even integer, we add 2:

x + 2 = 4 + 2 = 6

So, the two consecutive even integers that satisfy the given condition are 4 and 6.