Question 325791
<pre>
sqrt(4x+5) = 2 + sqrt(2x-1)
===========================
I wonder why the other person chose to write a novel for this problem!

         {{{sqrt(4x + 5) = 2 + sqrt(2x - 1)}}}, with {{{x>=1/2}}}
       {{{(sqrt(4x + 5))^2 = (2 + sqrt(2x - 1))^2}}} --- Squaring both sides of equation
           {{{4x + 5 = 4 + 4sqrt(2x - 1) + (2x - 1)}}}
           {{{4x + 5 = 2sqrt(2x - 1) + 2x + 3}}}
    {{{4x - 2x + 5 - 3 = 4sqrt(2x - 1)}}}
           {{{2x + 2 = 4sqrt(2x - 1)}}}
          {{{(2x + 2)/2 = (4sqrt(2x - 1))/2}}}
            {{{x + 1 = 2sqrt(2x - 1)}}}
          {{{(x + 1)^2 = (2sqrt(2x - 1))^2}}} ---- Squaring both sides of equation
       {{{x^2 + 2x + 1 = 4(2x - 1)}}}
       {{{x^2 + 2x + 1 = 8x - 4}}}
{{{x^2 + 2x - 8x + 1 + 4 = 0}}}
       {{{x^2 - 6x + 5 = 0}}}
  (x - 5)(x - 1) = 0
   x - 5 = 0      OR       x - 1 = 0
      <font color = red><font size = 4 ><b>x = 5</font></font></b>      OR          <font color = red><font size = 4 ><b>x = 1</font></font></b>
Both solutions are ACCEPTABLE, as {{{matrix(1,4, 5 >= 1/2, ",", and, 1 >= 1/2)}}}</pre