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?
The three integers are 9, 10, and 11.
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#
Since the problem states that the integers must be positive,
This means our three integers are
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#