SOLUTION: solve the following radical equation: {{{sqrt ( 3x+1 ) - sqrt ( 2x-10 )= 4 }}}

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Question 569282: solve the following radical equation: sqrt+%28+3x%2B1+%29+-+sqrt+%28+2x-10+%29=+4+
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
               sqrt+%28+3x%2B1+%29+-+sqrt+%28+2x-10+%29=+4+

Isolate one radical term on one side:

                  sqrt+%28+3x%2B1+%29+-+4 = sqrt+%28+2x-10+%29+

                %28sqrt+%28+3x%2B1+%29+-+4%29%5E2 = %28sqrt+%28+2x-10+%29%29%5E2+


    %28sqrt%283x%2B1%29%29%5E2 - 8sqrt%283x%2B1%29 + 16 = 2x - 10

    3x + 1 - 8sqrt%283x%2B1%29 + 16 = 2x - 10

        3x - 8sqrt%283x%2B1%29 + 17 = 2x - 10

Isolate the radical term on one side:

                  x + 27 = 8sqrt%283x%2B1%29

Square both sides:

               (x + 27)² = 8²%28sqrt%283x%2B1%29%29%5E2

          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+%28+3x%2B1+%29+-+sqrt+%28+2x-10+%29=+4+
sqrt+%28+3%2A5%2B1+%29+-+sqrt+%28+2%2A5-10+%29=+4+
sqrt+%28+15%2B1+%29+-+sqrt+%28+10-10+%29=+4+
sqrt+%2816%29+-+sqrt+%280+%29=+4+
4 - 0 = 4
    4 = 4

So x = 5 is a solution.

Checking x = 133 in the original:

sqrt+%28+3x%2B1+%29+-+sqrt+%28+2x-10+%29=+4+
sqrt+%28+3%2A133%2B1+%29+-+sqrt+%28+2%2A133-10+%29=+4+
sqrt+%28+399%2B1+%29+-+sqrt+%28+266-10+%29=+4+
sqrt+%28400%29+-+sqrt+%28256+%29=+4+
20 - 16 = 4
      4 = 4

So x = 133 is also a solution.

Edwin