Question 157511
{{{sqrt(2x+3)-sqrt(x+1)=1}}}
<pre><font size = 4 color = "indigo"><b>
Isolate one of the square root terms on one side
of the equation.  I will isolate the first term, {{{sqrt(2x+3)}}}
by adding {{{sqrt(x+1)}}} to both sides:

{{{sqrt(2x+3)=1-sqrt(x+1)}}}

Square both sides:

{{{(sqrt(2x+3))^2=(1-sqrt(x+1))^2}}}

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 {{{(1-sqrt(x+1))^2}}} as {{{(1-sqrt(x+1))(1-sqrt(x+1))}}}
and use FOIL:

{{{2x+3=(1-sqrt(x+1))(1-sqrt(x+1))}}}

{{{2x+3=1-sqrt(x+1)-sqrt(x+1)+x+1}}}

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

{{{2x+3=1-2sqrt(x+1)+x+1}}}

Simplify

{{{2x+3=2-2sqrt(x+1)+x}}}

Isolate the radical term on the left:

{{{2sqrt(x+1)=2+x-2x-3}}}

Simplify:

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

Square both sides:

{{{(2sqrt(x+1))^2=(-1-x)^2}}}

{{{2^2(sqrt(x+1))^2=(-1-x)(-1-x)}}}

{{{4(x+1)=1+x+x+x^2}}}

{{{4x+4=1+2x+x^2}}}

Get 0 on the left:

{{{0=1+2x+x^2-4x-4}}}

{{{0=x^2-2x-3}}}

Switch sides:

{{{x^2-2x-3=0}}}

{{{(x-3)(x+1)=0}}}

Set each factor = 0

{{{x-3=0}}} gives {{{x=3}}}

{{{x+1=0}}} gives {{{x=-1}}}

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

Checking {{{x=3}}} in original:

{{{sqrt(2x+3)-sqrt(x+1)=1}}}

{{{sqrt(2(3)+3)-sqrt(3+1)=1}}}

{{{sqrt(6+3)-sqrt(3+1)=1}}}

{{{sqrt(9)-sqrt(4)=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(2x+3)-sqrt(x+1)=1}}}

{{{sqrt(2(-1)+3)-sqrt(-1+1)=1}}}

{{{sqrt(-2+3)-sqrt(0)=1}}}

{{{sqrt(1)-0=1}}}

{{{1-0=1}}}

{{{1=1}}}

Therefore {{{x=-1}}} is also a solution.

Edwin</pre>