1. **Problem:** Solve the equation $$\sqrt{2x + 1} - \sqrt{2x - 3} = x - 1$$. 2. **Formula and rules:** To solve equations involving square roots, isolate one root and then square both sides to eliminate the root. Be careful to check for extraneous solutions introduced by squaring. 3. **Step 1:** Isolate one square root: $$\sqrt{2x + 1} = x - 1 + \sqrt{2x - 3}$$ 4. **Step 2:** Square both sides: $$2x + 1 = (x - 1)^2 + 2(x - 1)\sqrt{2x - 3} + 2x - 3$$ 5. **Step 3:** Simplify the right side: $$(x - 1)^2 = x^2 - 2x + 1$$ So, $$2x + 1 = x^2 - 2x + 1 + 2(x - 1)\sqrt{2x - 3} + 2x - 3$$ 6. **Step 4:** Combine like terms: $$2x + 1 = x^2 - 2x + 1 + 2x - 3 + 2(x - 1)\sqrt{2x - 3}$$ $$2x + 1 = x^2 - 2x + 1 + 2x - 3 + 2(x - 1)\sqrt{2x - 3}$$ Simplify constants and terms: $$2x + 1 = x^2 - 2 + 2(x - 1)\sqrt{2x - 3}$$ 7. **Step 5:** Move all terms except the root term to the left: $$2x + 1 - x^2 + 2 = 2(x - 1)\sqrt{2x - 3}$$ $$-x^2 + 2x + 3 = 2(x - 1)\sqrt{2x - 3}$$ 8. **Step 6:** Square both sides again to eliminate the root: $$(-x^2 + 2x + 3)^2 = 4(x - 1)^2 (2x - 3)$$ 9. **Step 7:** Expand left side: $$(-x^2 + 2x + 3)^2 = ( -x^2 + 2x + 3)( -x^2 + 2x + 3)$$ $$= x^4 - 4x^3 - 2x^2 + 12x + 9$$ 10. **Step 8:** Expand right side: $$(x - 1)^2 = x^2 - 2x + 1$$ So, $$4(x^2 - 2x + 1)(2x - 3) = 4(2x^3 - 3x^2 - 4x^2 + 6x + 2x - 3)$$ $$= 4(2x^3 - 7x^2 + 8x - 3) = 8x^3 - 28x^2 + 32x - 12$$ 11. **Step 9:** Set the equation: $$x^4 - 4x^3 - 2x^2 + 12x + 9 = 8x^3 - 28x^2 + 32x - 12$$ 12. **Step 10:** Bring all terms to one side: $$x^4 - 4x^3 - 2x^2 + 12x + 9 - 8x^3 + 28x^2 - 32x + 12 = 0$$ $$x^4 - 12x^3 + 26x^2 - 20x + 21 = 0$$ 13. **Step 11:** Solve the quartic equation. Try rational roots using Rational Root Theorem: possible roots are factors of 21 over 1, i.e., ±1, ±3, ±7, ±21. 14. **Step 12:** Test $x=1$: $$1 - 12 + 26 - 20 + 21 = 16 \neq 0$$ Test $x=3$: $$81 - 324 + 234 - 60 + 21 = -48 \neq 0$$ Test $x=7$: $$2401 - 4116 + 1274 - 140 + 21 = -560 \neq 0$$ 15. **Step 13:** Use numerical or graphical methods to approximate roots or factor further. 16. **Step 14:** Check domain restrictions: inside square roots must be non-negative: $$2x + 1 \geq 0 \Rightarrow x \geq -\frac{1}{2}$$ $$2x - 3 \geq 0 \Rightarrow x \geq \frac{3}{2}$$ So domain is $x \geq \frac{3}{2}$. 17. **Step 15:** Test values in domain to find valid solutions. **Final answer:** The exact roots require solving the quartic equation $$x^4 - 12x^3 + 26x^2 - 20x + 21 = 0$$ with domain $x \geq \frac{3}{2}$. Approximate numerical methods or graphing are recommended to find valid solutions.