Question 915375
{{{n}}}= the first of the 3 consecutive even integers.
So,
{{{n+2}}}= the second of the 3 consecutive even integers;
{{{n^2}}}= the square of the first integer, and
{{{4+16*(n+2)}}}= four more than sixteen times the second integer.
The problem says that
{{{n^2=4+16*(n+2)}}}
so we work from that equation.
{{{n^2=4+16*(n+2)}}}
{{{n^2=4+16*n+16*2}}}
{{{n^2=4+16n+32}}}
{{{n^2=16n+36}}}
{{{n^2-16n=36}}}
At this point, we can solve the quadratic equation by "completing the square":
{{{n^2-16n=36)}}}
{{{n^2-16n+64=36+64}}}
{{{n^2-16n+64=100}}}
{{{(n-8)^2=100}}}
{{{(n-8)^2=10^2}}} ---> {{{system(n-8=10,"or",n-8=-10)}}} ---> {{{system(n=10+8=18,"or",n=-10+8=-2)}}}
Since by "even integers" we usually understand, just positive integers divisible by 2, the only acceptable answer is
{{{highlight(n=18)}}} .
So the three consecutive even integers are {{{highlight(18)}}} , {{{highlight(20)}}} , and {{{highlight(22)}}} .