Question 519668
<pre>
how do you solve the following:

{{{sqrt(x+4)}}} + {{{sqrt(2x-1)}}} = {{{sqrt(7x+1)}}}
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One of the persons who responded states that {{{x = - 1/2}}}, but this is WRONG!! x CANNOT equal {{{- 1/2}}}.

Of the 3 RADICAL expressions, 2x - 1 is the SMALLEST. So, {{{2x - 1 >= 0}}}, and {{{2x >= 1}}} ===> {{{x >= 1/2}}}.
  So, we now get: {{{sqrt(x + 4)}}} + {{{sqrt(2x - 1)}}} = {{{sqrt(7x + 1)}}}, with {{{x >= 1/2}}}
                  {{{sqrt(x + 4)}}} + {{{sqrt(2x - 1)}}} = {{{sqrt(7x + 1)}}}
                  {{{(sqrt(x + 4) + sqrt(2x - 1))^2}}} = {{{(sqrt(7x + 1))^2}}} ----- Squaring both sides
{{{(sqrt(x + 4))^2 + 2sqrt(x + 4)sqrt(2x - 1) + (sqrt(2x - 1))^2}}} = 7x + 1
     {{{x + 4 + 2sqrt((x + 4)(2x - 1)) + 2x - 1}}} = 7x + 1
              {{{3x + 3 + 2sqrt(2x^2 + 7x - 4)}}} = 7x + 1
                     {{{2sqrt(2x^2 + 7x - 4)}}} = 7x + 1 - 3x - 3
                     {{{2sqrt(2x^2 + 7x - 4)}}} = 4x - 2
                  {{{(2sqrt(2x^2 + 7x - 4))^2}}} = {{{(4x - 2)^2}}} ------ Squaring both sides
                       {{{4(2x^2 + 7x - 4) = 16x^2 - 16x + 4}}}
                         {{{8x^2 + 28x - 16 = 16x^2 - 16x + 4}}}
                                    {{{0 = 16x^2 - 16x + 4 - 8x^2 - 28x + 16}}}
                                    {{{0 = 8x^2 - 44x + 20}}}
                                    {{{0 = 4(2x^2 - 11x + 5)}}}
                                    {{{0 = 2x^2 - 11x + 5}}}
                                    {{{0 = 2x^2 - 10x - x + 5}}}
                                    {{{0 = 2x(x - 5) - 1(x - 5)}}}
                                    0 = (2x - 1)(x - 5) 
                                    0 = 2x - 1      OR     0 = x - 5
                                    1 = 2x          OR    <font color = red><font size = 4><b>5 = x</font></font></b>
                                   {{{highlight(1/2 = x)}}}
As {{{x = 1/2 >= 1/2}}} and {{{x = 5 >= 1/2}}}, both solutions for x are VALID! </pre>