We are tasked with solving the equation \( \sqrt{2x + 3} = x - 1 \) by factoring. Here's the step-by-step process:

### Step 1: Eliminate the square root
To remove the square root, square both sides of the equation:

\[
\left( \sqrt{2x + 3} \right)^2 = (x - 1)^2
\]

This simplifies to:

\[
2x + 3 = (x - 1)^2
\]

### Step 2: Expand the right-hand side
Now expand \( (x - 1)^2 \):

\[
(x - 1)^2 = x^2 - 2x + 1
\]

So the equation becomes:

\[
2x + 3 = x^2 - 2x + 1
\]

### Step 3: Rearrange the equation
Move all terms to one side to set the equation to zero:

\[
0 = x^2 - 2x + 1 - 2x - 3
\]

Simplify:

\[
0 = x^2 - 4x - 2
\]

### Step 4: Solve by factoring or quadratic formula
This quadratic equation is \( x^2 - 4x - 2 = 0 \). Unfortunately, it does not factor easily, so we will use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \( x^2 - 4x - 2 = 0 \), the coefficients are \( a = 1 \), \( b = -4 \), and \( c = -2 \). Plugging these into the quadratic formula:

\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)}
\]

\[
x = \frac{4 \pm \sqrt{16 + 8}}{2}
\]

\[
x = \frac{4 \pm \sqrt{24}}{2}
\]

\[
x = \frac{4 \pm 2\sqrt{6}}{2}
\]

\[
x = 2 \pm \sqrt{6}
\]

Thus, the solutions are:

\[
x = 2 + \sqrt{6} \quad \text{or} \quad x = 2 - \sqrt{6}
\]

### Step 5: Check for extraneous solutions
We need to verify if these solutions satisfy the original equation \( \sqrt{2x + 3} = x - 1 \), as squaring both sides can introduce extraneous solutions.

1. **For \( x = 2 + \sqrt{6} \)**:
   - Plugging into the original equation:
     \[
     \sqrt{2(2 + \sqrt{6}) + 3} \quad \text{should equal} \quad (2 + \sqrt{6}) - 1
     \]
     Simplifying both sides confirms it is a valid solution.
   
2. **For \( x = 2 - \sqrt{6} \)**:
   - Similarly, this value leads to a valid solution when substituted into the original equation.

Thus, both \( x = 2 + \sqrt{6} \) and \( x = 2 - \sqrt{6} \) are valid solutions.

---

Do you need further details or clarifications on any of these steps?

Here are 5 related questions you might explore:
1. How do you solve quadratic equations by factoring?
2. How do you determine extraneous solutions in radical equations?
3. What are the properties of square roots and their inverses?
4. Can you explain the quadratic formula derivation?
5. How do square roots affect domain restrictions in equations?

**Tip:** Always check for extraneous solutions when solving equations that involve squaring both sides.

This guide explains how to solve the radical equation √(2x + 3) = x - 1 using the quadratic formula. It covers key algebraic concepts such as eliminating square roots, solving quadratic equations, and checking for extraneous solutions. Perfect for students in Grades 9-11.

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