\sqrt{2x + 3} – \sqrt{x + 1}= 1

↑

\fbox{Isolate one of the radicals on one side of the equation.}

Step 1   \sqrt{2x + 3} = 1 +\sqrt{x + 1}                         Isolate \sqrt{2x + 3}.

Step 2  (\sqrt{2x + 3})² = (1 + \sqrt{x + 1})²                    Square each side.

2x + 3  =  1 +  2\sqrt{x + 1} +(x + 1)          Be careful:

↑                                         (a + b)² = a² + 2ab + b²

\fbox{Don’t forget this term when squaring.}

Step 1           x + 1 = 2\sqrt{x + 1}                          Isolate the remaining radical.

Step 2           (x + 1)² =(2\sqrt{x + 1})²                    Square again.

x² + 2x + 1 = 4(x + 1)←(ab)² = a²b²    Apply the exponents.

x² + 2x + 1 = 4x + 4                             Distributive property

Step 3            x² – 2x – 3 = 0                                 Write in standard form.

(x – 3)(x + 1) = 0                                            Factor.

x – 3 = 0   or     x + 1 = 0                               Zero-factor property

x = 3         or     x = -1                                     Proposed solutions

Step 4

CHECK  \sqrt{2x + 3} – \sqrt{x + 1} = 1         Original equation

\left. \begin{matrix}\sqrt{2(3)+3}-\sqrt{3+1} ≟ 1&\text{Let}   x =3.\\ \sqrt{9}-\sqrt{4}≟ 1\\ 3-2≟ 1\\1=1 ✓& \text{True}\end{matrix}\right|\begin{matrix}\sqrt{2(-1)+3}-\sqrt{-1+1} ≟ 1&\text{Let}   x = -1.\\\sqrt{1}-\sqrt{0}≟ 1\\ 1-0≟ 1\\1=1 ✓& \text{ True}\end{matrix}

Both 3 and -1 are solutions of the original equation, so { -1, 3} is the solution set.