Question 597322
<pre>
Please help me solve this equation {{{ sqrt (3x+4)=2+ sqrt (x)}}}
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There ISN'T just 1 value for x, as the other person indicates!

    {{{sqrt(3x + 4) = 2 + sqrt(x)}}}
 {{{(sqrt(3x + 4))^2 = (2 + sqrt(x))^2}}} ---- Squaring each side
     {{{3x + 4 = 4 + 4sqrt(x) + x}}}
{{{3x + 4 - 4 - x = 4sqrt(x)}}}
        {{{2x = 4sqrt(x)}}}
      {{{(2x)^2 = (4sqrt(x))^2}}} ------ Squaring each side
       {{{4x^2 = 16(x)}}} 
  {{{4x^2 - 16x = 0}}}
   {{{4x(x - 4) = 0}}} 
        4x = 0      OR     x - 4 = 0
        <font color = red><font size = 4><b>x = 0</font></font></b>      OR        <font color = red><font size = 4><b>x = 4</font></font></b>

Both solutions are VALID!!</pre>