Solve the equation sqrt(2x + 1) = x - 1.

Step-by-step solution

1. 1

   Fix the domain before squaring. The left side is a principal square root, so it is $\ge 0$; the right side must therefore also be $\ge 0$, giving $x \ge 1$. The radicand needs $2x + 1 \ge 0$, i.e. $x \ge -\tfrac12$, which is weaker. The binding condition is

   $x \ge 1$

   Squaring is not reversible, so this condition is the only thing that will separate real solutions from artefacts later.

2. 2

   Square both sides.

   $2x + 1 = (x-1)^2 = x^2 - 2x + 1$

3. 3

   Collect and factor.

   $0 = x^2 - 4x = x(x-4) \quad\Longrightarrow\quad x = 0 \ \text{or}\ x = 4$

   The constant $1$ cancelled from both sides, which is why the quadratic factors so cleanly.

4. 4

   Test both candidates against the domain and the original equation. $x = 0$ fails $x \ge 1$; substituting anyway gives $\sqrt{1} = 1$ on the left but $0 - 1 = -1$ on the right, so it is extraneous — it solves the squared equation only. For $x = 4$: $\sqrt{2(4)+1} = \sqrt{9} = 3$ and $4 - 1 = 3$. It works.

5. 5

   State the solution.

   $x = 4$

   Geometrically, the curve $y = \sqrt{2x+1}$ and the line $y = x - 1$ cross exactly once, at $(4,3)$; the point $x = 0$ is where the line meets the reflected branch $y = -\sqrt{2x+1}$, which squaring silently folded in.