1. **State the problem:** Solve the equation $$x - 1 = \sqrt{x + 1}$$ for $x$. 2. **Recall the formula and rules:** To solve equations involving square roots, we isolate the square root term and then square both sides to eliminate the root. Remember to check for extraneous solutions after squaring. 3. **Isolate the square root:** The equation is already isolated: $$x - 1 = \sqrt{x + 1}$$ 4. **Square both sides:** $$ (x - 1)^2 = (\sqrt{x + 1})^2 $$ $$ (x - 1)^2 = x + 1 $$ 5. **Expand the left side:** $$ (x - 1)^2 = x^2 - 2x + 1 $$ 6. **Set up the quadratic equation:** $$ x^2 - 2x + 1 = x + 1 $$ 7. **Bring all terms to one side:** $$ x^2 - 2x + 1 - x - 1 = 0 $$ $$ x^2 - 3x = 0 $$ 8. **Factor the quadratic:** $$ x(x - 3) = 0 $$ 9. **Solve for $x$:** $$ x = 0 \quad \text{or} \quad x = 3 $$ 10. **Check for extraneous solutions:** - For $x=0$: $$ 0 - 1 = -1 $$ $$ \sqrt{0 + 1} = 1 $$ Left side $\neq$ right side, so $x=0$ is extraneous. - For $x=3$: $$ 3 - 1 = 2 $$ $$ \sqrt{3 + 1} = \sqrt{4} = 2 $$ Left side $=$ right side, so $x=3$ is a valid solution. **Final answer:** $x=3$ (option c).