Question 331424
Solve the quadratic equation by using the COMPLETING THE SQUARE method and 
applying the square root property.
;
I think the problem should be:
{{{n^2-(2/3)n = 1/9}}}
:
{{{n^2-(2/3)n + ____ = 1/9}}}
:
{{{(1/3)^2}}} = {{{1/9}}}completes the square
{{{n^2-(2/3)n + 1/9}}} = {{{1/9 + 1/9}}}
{{{n^2-(2/3)n + 1/9}}} = {{{2/9}}}
{{{(n-1/3)^2}}} = {{{2/9}}}
Find the square root of both sides
{{{n-1/3}}} = +/-{{{sqrt(2/9)}}}
Extract 1/3 from 1/9
{{{n-1/3}}} = +/-{{{sqrt(2)/3}}}
n = {{{1/3}}}+/-{{{sqrt(2)/3}}}
Two solutions
n = {{{(1 + sqrt(2))/3}}}
and
n = {{{(1 - sqrt(2))/3}}}
:
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One way to check this problem: 
Use a cal, find that the approx decimal value for {{{(1 + sqrt(2))/3}}} = .8047
Substitute this value for n in the original problem find the value
.8047^2 -{{{2/3}}}(.8047) = .111 which is the decimal value of 1/9, ,1111
confirms our solution, you can do the same with {{{(1 - sqrt(2))/3}}}