Question 661043
<pre>
 ___        ___
&#8730;x-3 + 1 = &#8730;x+2

You have one of the radical terms already 
isolated on the right, so we can go ahead and
square both sides:

  ___           ___
(&#8730;x-3 + 1)² = (&#8730;x+2)²

It's easy to square the right side by just removing 
the radical and the exponent and just get x+2.

  ___          
(&#8730;x-3 + 1)² = x+2

But to square the left side is not so easy because 
it has two terms:
  ___           ___       ___
(&#8730;x-3 + 1)² = (&#8730;x-3 + 1)(&#8730;x-3 + 1) 

So we have to use FOIL: <font color = "red">
 ___ ___    ___    ___
&#8730;x-3&#8730;x-3 + &#8730;x-3 + &#8730;x-3 + 1
  ___       ___
(&#8730;x-3)² + 2&#8730;x-3 + 1
            ___
  x-3   + 2&#8730;x-3 + 1
            ___
  x - 2 + 2&#8730;x-3
</font>
So now our equation:

    ___          
  (&#8730;x-3 + 1)² = x+2

becomes

          ___
x - 2 + 2&#8730;x-3 = x+2

We isolate the radical term on the left 
         ___
       2&#8730;x-3 = 4

We can divide both sides by 2
         ___
        &#8730;x-3 = 2

We square both sides:

        ___
      (&#8730;x-3)² = (2)²

          x-3 = 4

            x = 7

That may or not be the solution.  So we
must always check, because these equations
may have bogus solutions (called "extraneous"),
So we must check in the ORIGINAL equation:

 ___        ___
&#8730;x-3 + 1 = &#8730;x+2
 ___        ___
&#8730;7-3 + 1 = &#8730;7+2
   _        _
  &#8730;4 + 1 = &#8730;9   

   2 + 1 = 3

       3 = 3

It checks, so the solution is x=7.

Edwin</pre>