Question

find three consecutive even integers such that the product of the first and second is 4 less than the square of the third.

Answer 1

If the smallest is x, then you want
x(x+2) = (x+4)^2 - 4
Now just solve for x

Answer 2

Three consecutive even integers are:

2 n , 2 n + 2 , 2 n + 4

Product of the first and second is 4 less than the square of the third means:

2 n ∙ ( 2 n + 2 ) = ( 2 n + 4 )² - 4

4 n² + 4 n = 4 n² + 2 ∙ 2 n ∙ 4 + 4 ² - 4

4 n² + 4 n = 4 n² + 16 n + 16 - 4

4 n² + 4 n = 4 n² + 16 n + 12

Subtract 4 n² to both sides

4 n = 16 n + 12

Subtract 16 n to both sides

- 12 n = 12

Divide both sides by - 12

n = - 1

Your numbers are:

2 n , 2 n + 2 , 2 n + 4

- 2 , 0 , 2

Proof:

Product of the first and second is 4 less than the square of the third.

( - 2 ) ∙ 0 = 0

2² = 4

( - 2 ) ∙ 0 = 2² - 4 = 4 - 4

Answer 3

n is the 1st integer

n(n + 2) = [(n + 4)^2] - 4 ... n^2 + 2n + 4 = n^2 + 8n + 16