Question

can someone please help with this :)
identify three consecutive positive integers such that nine times the largest is exactly four less than the square of the middle value

Answer 1

Answer:

9, 11, 13

Step-by-step explanation:

The integers are x, (x+2), and (x+5).

9(x + 4) = (x+2)^2 - 4

9x + 36 = (x+2)(x+2) - 4

9x + 36 = x^2 + 4x + 4 - 4

9x + 36 = x^2 + 4x

Subtract x^2 + 4x from both sides

-x^2 + 9x - 4x + 36 = 0

-x^2 + 5x + 36 = 0

Divide each side by -1

x^2 - 5x - 36 = 0

Factor

(x+4)(x-9) = 0

Solve

-4 + 4 = 0 and 9 - 9 = 0

Since it is positive and odd exponents, x will not be 4, it will be 9.

x = 9     9 + 2 = 11     9 + 4 = 13

9, 11, 13

-------------Check-------------

9(x + 4) = (x + 2)^2 - 4

9(9 + 4) = (9 + 2)^2 - 4

81 + 36 = (9 + 2)(9 + 2) - 4

117 = 81 + 18 + 18 + 4 - 4

117 = 117