SOLUTION: {{{sqrt(2x+3)-sqrt(x+1)=1}}} can someone explain this problem to me?

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Question 157511: sqrt%282x%2B3%29-sqrt%28x%2B1%29=1 can someone explain this problem to me?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x%2B3%29-sqrt%28x%2B1%29=1

Isolate one of the square root terms on one side
of the equation.  I will isolate the first term, sqrt%282x%2B3%29
by adding sqrt%28x%2B1%29 to both sides:

sqrt%282x%2B3%29=1-sqrt%28x%2B1%29

Square both sides:

%28sqrt%282x%2B3%29%29%5E2=%281-sqrt%28x%2B1%29%29%5E2

The left side amounts to squaring a square root,
which just amounts to taking away the radical.

The right side is not so easy.  Write %281-sqrt%28x%2B1%29%29%5E2 as %281-sqrt%28x%2B1%29%29%281-sqrt%28x%2B1%29%29
and use FOIL:

2x%2B3=%281-sqrt%28x%2B1%29%29%281-sqrt%28x%2B1%29%29

2x%2B3=1-sqrt%28x%2B1%29-sqrt%28x%2B1%29%2Bx%2B1

Combine the two middle terms, which are like terms, 
on the right:

2x%2B3=1-2sqrt%28x%2B1%29%2Bx%2B1

Simplify

2x%2B3=2-2sqrt%28x%2B1%29%2Bx

Isolate the radical term on the left:

2sqrt%28x%2B1%29=2%2Bx-2x-3

Simplify:

2sqrt%28x%2B1%29=-1-x

Square both sides:

%282sqrt%28x%2B1%29%29%5E2=%28-1-x%29%5E2

2%5E2%28sqrt%28x%2B1%29%29%5E2=%28-1-x%29%28-1-x%29

4%28x%2B1%29=1%2Bx%2Bx%2Bx%5E2

4x%2B4=1%2B2x%2Bx%5E2

Get 0 on the left:

0=1%2B2x%2Bx%5E2-4x-4

0=x%5E2-2x-3

Switch sides:

x%5E2-2x-3=0

%28x-3%29%28x%2B1%29=0

Set each factor = 0

x-3=0 gives x=3

x%2B1=0 gives x=-1

But we must check these because
many times we get extraneous solutions:

Checking x=3 in original:

sqrt%282x%2B3%29-sqrt%28x%2B1%29=1

sqrt%282%283%29%2B3%29-sqrt%283%2B1%29=1

sqrt%286%2B3%29-sqrt%283%2B1%29=1

sqrt%289%29-sqrt%284%29=1

3-2=1

1=1

Therefore x=3 is a solution.

But we must check these because
many times we get extraneous solutions:

Checking x=-1 in original:

sqrt%282x%2B3%29-sqrt%28x%2B1%29=1

sqrt%282%28-1%29%2B3%29-sqrt%28-1%2B1%29=1

sqrt%28-2%2B3%29-sqrt%280%29=1

sqrt%281%29-0=1

1-0=1

1=1

Therefore x=-1 is also a solution.

Edwin