Find three consecutive positive integers such that the square of the first increased by twice the second is three less than four times the third.

  3, 4, 5

Step-by-step explanation:

I like to let x represent the middle integer. Then the first is x-1 and the third is x+1. The given relation is ...

  (x-1)^2 +2x = 4(x+1) -3

We can put this equation into standard form and solve.

  x^2 -2x +1 +2x = 4x +4 -3

  x^2 -4x = 0 . . . . . subtract 4x+1

  x(x -4) = 0

So, x=0 is a solution to the equation, but an extraneous solution with respect to the problem.

x = 4 is also a solution to the equation, indicating the integers are 3, 4, 5.

Check

3^2 +2·4 = 4·5 -3

17 = 17 . . . . answer checks OK

i think c. but don't trust it.

step-by-step explanation: