Question 30998

{{{ sqrt(3x-5) - sqrt(x+7) = 2 }}}
{{{ sqrt(3x-5) = 2 + sqrt(x+7) }}}
{{{ (sqrt(3x-5))^2 = (2 + sqrt(x+7))^2 }}}
{{{ 3x-5 = (2 + sqrt(x+7))(2 + sqrt(x+7)) }}}
{{{ 3x-5 = 4 + 2sqrt(x+7) + 2sqrt(x+7) + sqrt(x+7)sqrt(x+7) }}}
{{{ 3x-5 = 4 + 4sqrt(x+7) + (x+7) }}}
{{{ 3x-5 = 11 + 4sqrt(x+7) + x }}}
{{{ 2x-16 = 4sqrt(x+7) }}}
{{{ (2x-16)/4 = sqrt(x+7) }}}
{{{ (x/2)-4 = sqrt(x+7) }}}
{{{ ((x/2)-4)^2 = (sqrt(x+7))^2 }}}
{{{ ((x/2)-4)((x/2)-4) = x+7 }}}
{{{ (x/2)^2 -4(x/2) - 4(x/2) + 16 = x+7 }}}
{{{ (x^2/4) -4x + 16 = x+7 }}}
{{{ (x^2/4) -5x + 16 = 7 }}}
{{{ (x^2/4) -5x + 9 = 0 }}}
{{{ x^2 -20x + 36 = 0 }}}
(x-18)(x-2) = 0


so x-18 = 0 or x-2=0
--> x=18 or x=2


CHECK:
when x=18: {{{ sqrt(3x-5) - sqrt(x+7) }}} becomes
{{{ sqrt(3(18)-5) - sqrt((18)+7) }}}
{{{ sqrt(54-5) - sqrt(18+7) }}}
{{{ sqrt(49) - sqrt(25) }}}
{{{ 7 - 5 }}}
--> 2


also, when x=2: {{{ sqrt(3x-5) - sqrt(x+7) }}} becomes
{{{ sqrt(3(2)-5) - sqrt(2+7) }}}
{{{ sqrt(6-5) - sqrt(2+7) }}}
{{{ sqrt(1) - sqrt(9) }}}


now...tricky: {{{sqrt(1)}}} is either +1 or -1. Similarly {{{sqrt(9)}}} is either +3 or -3. So, we get


-1 - (-3)
-1 + 3
--> 2


jon.