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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.

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Answer:

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

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

  • The calculation is as follows:

Let us assume the first even integer = n

Let us assume the second even integer = n + 2

Let us assume the third even integer = n + 4

Now

[tex](n^2) + (n + 2)^2 + 4 = 2(n + 4)^2\\\\n^2 + (n + 2)(n + 2) + 4 = 2(n + 4)(n + 4)\\\\n^2 + n^2 + 2n + 2n + 4 + 4 = 2(n^2 + 4n + 4n + 16)\\\\2n^2 + 4n + 8 = 2(n^2 + 8n + 16)\\\\2n^2 + 4n + 8 = 2n^2 + 16n + 32\\\\2n^2 - 2n^2 + 4n - 16n = 32 - 8\\\\- 12n = 24[/tex]

n = -2

Now  

Second integer = n + 2

= -2 + 2 = 0

And,  

Third integer = n + 4  

= -2 + 4

= 2

Learn more: brainly.com/question/17429689

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