SOLUTION: Hello! Can you please solve this question right away since I need this immediately for tomorrow: Instruction(s): Find all real solutions in each equation. √(x-3)+1=&#87

 ___        ___
√x-3 + 1 = √x+2

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

  ___           ___
(√x-3 + 1)² = (√x+2)²

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

  ___          
(√x-3 + 1)² = x+2

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

So we have to use FOIL: 
 ___ ___    ___    ___
√x-3√x-3 + √x-3 + √x-3 + 1
  ___       ___
(√x-3)² + 2√x-3 + 1
            ___
  x-3   + 2√x-3 + 1
            ___
  x - 2 + 2√x-3

So now our equation:

    ___          
  (√x-3 + 1)² = x+2

becomes

          ___
x - 2 + 2√x-3 = x+2

We isolate the radical term on the left 
         ___
       2√x-3 = 4

We can divide both sides by 2
         ___
        √x-3 = 2

We square both sides:

        ___
      (√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:

 ___        ___
√x-3 + 1 = √x+2
 ___        ___
√7-3 + 1 = √7+2
   _        _
  √4 + 1 = √9   

   2 + 1 = 3

       3 = 3

It checks, so the solution is x=7.

Edwin