Question 519668
<pre>

The other tutor's solution is incorrect

{{{sqrt(x+4)}}} + {{{sqrt(2x-1)}}} = {{{sqrt(7x+1)}}}

Since one of the radicals is isolated, we square both sides:

To make things a little easier, 

let A = {{{sqrt(x+4)}}}
let B = {{{sqrt(2x-1)}}}
let C = {{{sqrt(7x+1)}}}

Then the above equation becomes:

         A + B = C

Squaring both sides:

      (A + B)² = C²

(A + B)(A + B) = C²

 A² + 2AB + B² = C² 

Since A = {{{sqrt(x+4)}}}, then A² = x+4
Since B = {{{sqrt(2x-1)}}}, then B² = 2x-1
let C = {{{sqrt(7x+1)}}}, then C² = 7x+1

So now we have

 x+4 + 2AB + 2x-1 = 7x+1

Simplify and solve for 2AB

 3x + 3 + 2AB = 7x + 1
          2AB = 4x - 2

Divide through by 2 since all coefficients are even:

           AB = 2x - 1

Square both sides again:

        (AB)² = (2x - 1)²

         A²B² = (2x - 1)(2x - 1)              

         A²B² = 4x² - 4x + 1

Now substitute A² = x+4  and B² = 2x-1 from above:

 (x+4)(2x-1) = 4x² - 4x + 1 

 2x² + 7x - 4 = 4x² - 4x + 1

 -2x² + 11x - 5 = 0

  2x² - 11x + 5 = 0

(x - 5)(2x - 1) = 0

x - 5 = 0;  2x - 1 = 0

    x = 5;      2x = 1

                 x = {{{1/2}}}

Check each solution as they may be extraneous: 
             
Checking x = 5

{{{sqrt(x+4)}}} + {{{sqrt(2x-1)}}} = {{{sqrt(7x+1)}}}
{{{sqrt(5+4)}}} + {{{sqrt(2(5)-1)}}} = {{{sqrt(7(5)+1)}}}
{{{sqrt(9)}}} + {{{sqrt(9)}}} = {{{sqrt(36)}}}
3 + 3 = 6
    6 = 6

That checks so x = 5 is a solution.

Checking x = {{{1/2}}}

{{{sqrt(x+4)}}} + {{{sqrt(2x-1)}}} = {{{sqrt(7x+1)}}}
{{{sqrt(1/2+4)}}} + {{{sqrt(2(1/2)-1)}}} = {{{sqrt(7(1/2)+1)}}}
{{{sqrt(1/2+8/2)}}} + {{{sqrt(1-1)}}} = {{{sqrt(7/2+1)}}}
{{{sqrt(9/2)}}} + {{{sqrt(0)}}} = {{{sqrt(7/2+2/2)}}}
{{{sqrt(9/2)}}} + 0 = {{{sqrt(9/2)}}}
{{{sqrt(9/2)}}} = {{{sqrt(9/2)}}}

That checks also, so x = {{{1/2}}} is also a solution.

Edwin</pre>