1. **State the problem:** Solve the equation and check for extraneous solutions. 2. **General approach:** When solving equations, especially those involving radicals, fractions, or variables on both sides, we must check for extraneous solutions by substituting back into the original equation. 3. **Example problem:** Suppose the equation is $\sqrt{x+3} = x - 1$. 4. **Step 1: Isolate the radical:** The radical is already isolated. 5. **Step 2: Square both sides to eliminate the square root:** $$\left(\sqrt{x+3}\right)^2 = (x - 1)^2$$ $$x + 3 = (x - 1)^2$$ 6. **Step 3: Expand the right side:** $$x + 3 = x^2 - 2x + 1$$ 7. **Step 4: Rearrange to form a quadratic equation:** $$0 = x^2 - 2x + 1 - x - 3$$ $$0 = x^2 - 3x - 2$$ 8. **Step 5: Factor the quadratic:** $$0 = (x - 2)(x - (-1)) = (x - 2)(x + 1)$$ 9. **Step 6: Solve for $x$:** $$x = 2 \quad \text{or} \quad x = -1$$ 10. **Step 7: Check for extraneous solutions by substituting back into the original equation:** - For $x=2$: $$\sqrt{2+3} = \sqrt{5} \approx 2.236$$ $$2 - 1 = 1$$ Not equal, so $x=2$ is extraneous. - For $x=-1$: $$\sqrt{-1+3} = \sqrt{2} \approx 1.414$$ $$-1 - 1 = -2$$ Not equal, so $x=-1$ is extraneous. 11. **Step 8: Conclusion:** Both solutions are extraneous, so the equation has no valid solutions. **Final answer:** No solution.