Question 569282
<pre>
               {{{sqrt ( 3x+1 ) - sqrt ( 2x-10 )= 4 }}}

Isolate one radical term on one side:

                  {{{sqrt ( 3x+1 ) - 4}}} = {{{sqrt ( 2x-10 ) }}}

                {{{(sqrt ( 3x+1 ) - 4)^2}}} = {{{(sqrt ( 2x-10 ))^2 }}}


    {{{(sqrt(3x+1))^2}}} - 8{{{sqrt(3x+1)}}} + 16 = 2x - 10

    3x + 1 - 8{{{sqrt(3x+1)}}} + 16 = 2x - 10

        3x - 8{{{sqrt(3x+1)}}} + 17 = 2x - 10

Isolate the radical term on one side:

                  x + 27 = 8{{{sqrt(3x+1)}}}

Square both sides:

               (x + 27)² = 8²{{{(sqrt(3x+1))^2}}}

          x² + 54x + 729 = 64(3x + 1)

          x² + 54x + 729 = 192x + 64

Get 0 on the right:

         x² - 138x + 665 = 0         

Factor:

        (x - 5)(x - 133) = 0

Use the zero factor principle:

    x - 5 = 0        x - 133 = 0 
        x = 5              x = 133

We much check all solution because squaring both sides
often but not always introduces extraneous answers.

Checking x = 5 in the original:

{{{sqrt ( 3x+1 ) - sqrt ( 2x-10 )= 4 }}}
{{{sqrt ( 3*5+1 ) - sqrt ( 2*5-10 )= 4 }}}
{{{sqrt ( 15+1 ) - sqrt ( 10-10 )= 4 }}}
{{{sqrt (16) - sqrt (0 )= 4 }}}
4 - 0 = 4
    4 = 4

So x = 5 is a solution.

Checking x = 133 in the original:

{{{sqrt ( 3x+1 ) - sqrt ( 2x-10 )= 4 }}}
{{{sqrt ( 3*133+1 ) - sqrt ( 2*133-10 )= 4 }}}
{{{sqrt ( 399+1 ) - sqrt ( 266-10 )= 4 }}}
{{{sqrt (400) - sqrt (256 )= 4 }}}
20 - 16 = 4
      4 = 4

So x = 133 is also a solution.

Edwin</pre>