Question

Find three consecutive even integers such that the square of the first is 80 less than the square of the third.

Answer 1

8, 10, 12

Answer 2

oh didnt read even....lemme change that

let x = first integer
x + 2 = second integer
x + 4 = 3rd integer

Answer 3

\[X^2+80=(x+4)^2\]

Answer 4

\[x^2+80=(x+4)^2\]\[x^2+80=(x+4)(X+4)\]\[x^2+80=x^2+8x+16\]\[8x=64\]\[x=8\]

Answer 5

\[x=8\]\[x+2=10\]\[x+4=12\]