If the smallest is x, then you want
x(x+2) = (x+4)^2 - 4
Now just solve for x
find three consecutive even integers such that the product of the first and second is 4 less than the square of the third.
3 answers
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
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
n is the 1st integer
n(n + 2) = [(n + 4)^2] - 4 ... n^2 + 2n + 4 = n^2 + 8n + 16
n(n + 2) = [(n + 4)^2] - 4 ... n^2 + 2n + 4 = n^2 + 8n + 16