Question

In 3 consecutive positive integers, the square of the 1st integer is equal to the sum of, 7 times the second integer, and the 3rd integer. How to find all 3 integers?

Answer 1

The three integers are 9, 10, and 11.

Explanation:

I assume that you mean the first integer squared is equal to 7 times the second integer PLUS the third integer. Otherwise, there is no integer solution to this problem.

Using the information from the problem:

"Square of the 1st integer is 7 times the second integer plus the third integer"

`n^2 = 7(n+1) + (n+2)`

`n^2 = 7n + 7 + n + 2`

`n^2 = 8n + 9`

`n^2 - 8n - 9 = 0`

`(n-9)(n+1) = 0`

`n = 9 " "or " "n=-1`

Since the problem states that the integers must be positive, `n=9`.

This means our three integers are `9, 10, and 11`.

Final Answer

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We can check this work by plugging the numbers we found back into our original problem:

"The square of the first integer equals 7 times the second integer plus the third integer"

`(9)^2 stackrel(color(red)?color(white)xx)(=) 7(10) + 11`

`81 stackrel(color(limegreen)sqrt()color(white)xx)(=) 81`