Find three consecutive even integers such that the sum of the squares of the first two is four less than twice the square of the third.

the three consecutive even integers are -2, 0, 2

Step-by-step explanation:

Let the first even integer = n

Let the second even integer = n + 2

Let the third even integer = n + 4

From the given statement, we form the equation below

(n²) + (n + 2)²  + 4 = 2(n + 4)²

Expand this equation;

n² + (n + 2)(n + 2) +  4 = 2(n + 4)(n + 4)

n² + n² + 2n + 2n +  4 +  4 = 2(n² + 4n + 4n + 16)

2n² + 4n + 8 = 2(n² + 8n + 16)

2n² + 4n  + 8 = 2n² + 16n + 32

collect like terms together;

2n² - 2n² + 4n - 16n = 32 - 8

- 12n = 24

-n = 24/12

-n = 2

n = -2

Second integer = n + 2

                          = -2 + 2 = 0

Third integer = n + 4

                      = -2 + 4

                      = 2

Therefore, the three consecutive even integers are -2, 0, 2

she gotta workk that a 2

step-by-step explanation: