Question 707329
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How would you solve this equation?

{{{ sqrt ( 2x ) = sqrt ( 3x+12 )-2 }}}
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The other person's response, that "You get -10+6i or -10-6i (complex results)", is WRONG11

{{{sqrt(2x) = sqrt(3x+12) - 2}}}
As the smaller radicand, 2x MUST be greater than or equal to 0, we get: {{{2x >= 0}}}. So, {{{x >= 0}}}.  
This gives us:      {{{sqrt(2x) = sqrt(3x+12) - 2}}}, with {{{x >= 0}}}
                         {{{(sqrt(2x))^2 = (sqrt(3x + 12) - 2)^2}}} ---- Squaring both sides
                               {{{2x = 3x + 12 - 4sqrt(3x + 12) + 4}}}
                    {{{4sqrt(3x + 12) = 3x + 12 + 4 - 2x}}}
                    {{{4sqrt(3x + 12) = x + 16}}}
                  {{{16(3x + 12) = x^2 + 32x + 256}}} ----- Squaring both sides
                     {{{48x + 192 = x^2 + 32x + 256}}}
{{{x^2 + 32x + 256 - 48x - 192 = 0}}}
                 {{{x^2 - 16x + 64 = 0}}}
                        {{{(x - 8)^2 = 0}}}
                          x - 8 = 0.
                               {{{highlight(x = 8)}}} <==== VALID, and ACCEPTABLE solution for x, in THIS case!!</font></font></font></b></pre>