Question 182132
the problem is radical 3+2n + radical 2-2n = 3,
:
{{{sqrt(3+2n)}}} + {{{sqrt(2-2n)}}} = 3
:
{{{sqrt(3+2n)}}} = 3 - {{{sqrt(2-2n)}}}
square both sides:
3 + 2n = (3 - {{{sqrt(2-2n)}}})^2
FOIL the right side
3 + 2n = 9 - 6{{{sqrt(2-2n)}}} + (2-2n)
:
3 + 2n = 9 + 2 - {{{6sqrt(2-2n)}}} - 2n
: 
3 + 2n = 11 - {{{6sqrt(2-2n)}}} - 2n
:
3 - 11 + 2n + 2n = -{{{6sqrt(2-2n)}}}
:
4n - 8 = -{{{6sqrt(2-2n)}}}
:
square both sides, FOIL the left side
16n^2 - 64n + 64 = 36(2-2n)
:
16n^2 - 64n + 64 = 72 - 72n
:
16n^2 - 64n + 72n + 64 - 72 = 0
:
16n^2 + 8n - 8 = 0
Simplify, divide by 8
2n^2 + n - 1 = 0
Factors to
(2n - 1)(n + 1) = 0
Two solutions:
n = {{{1/2}}}
n = -1
:
:
Check both solution in original equation:
n = .5
{{{sqrt(3+2(.5))}}} + {{{sqrt(2-2(.5))}}} = 3
{{{sqrt(3+1)}}} + {{{sqrt(2-1)}}} = 3
{{{sqrt(4)}}} + {{{sqrt(1)}}} = 3; a good solution
:
n = -1
{{{sqrt(3-2)}}} + {{{sqrt(2-(-2))}}} = 3
{{{sqrt(1)}}} + {{{sqrt(4)}}} = 3; also a good solution