1. Stating the problem: Solve the equation $$\sqrt{x-2} - \sqrt{x-4} = \sqrt{x+3}$$ for $x$. 2. Important domain considerations: The expressions under the square roots must be non-negative. - For $\sqrt{x-2}$, require $x-2 \geq 0 \Rightarrow x \geq 2$. - For $\sqrt{x-4}$, require $x-4 \geq 0 \Rightarrow x \geq 4$. - For $\sqrt{x+3}$, require $x+3 \geq 0 \Rightarrow x \geq -3$. Combining these, the domain is $x \geq 4$. 3. Isolate one square root to prepare for squaring: $$\sqrt{x-2} = \sqrt{x-4} + \sqrt{x+3}$$ 4. Square both sides: $$\left(\sqrt{x-2}\right)^2 = \left(\sqrt{x-4} + \sqrt{x+3}\right)^2$$ $$x-2 = (x-4) + (x+3) + 2\sqrt{(x-4)(x+3)}$$ 5. Simplify the right side: $$x-2 = 2x -1 + 2\sqrt{(x-4)(x+3)}$$ 6. Rearrange to isolate the square root term: $$x-2 - 2x + 1 = 2\sqrt{(x-4)(x+3)}$$ $$-x -1 = 2\sqrt{(x-4)(x+3)}$$ 7. Divide both sides by 2: $$\frac{-x -1}{2} = \sqrt{(x-4)(x+3)}$$ 8. Square both sides again: $$\left(\frac{-x -1}{2}\right)^2 = (x-4)(x+3)$$ $$\frac{(x+1)^2}{4} = x^2 - x -12$$ 9. Multiply both sides by 4: $$(x+1)^2 = 4x^2 - 4x - 48$$ 10. Expand left side: $$x^2 + 2x + 1 = 4x^2 - 4x - 48$$ 11. Bring all terms to one side: $$0 = 4x^2 - 4x - 48 - x^2 - 2x - 1$$ $$0 = 3x^2 - 6x - 49$$ 12. Solve quadratic equation: $$3x^2 - 6x - 49 = 0$$ Use quadratic formula: $$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot (-49)}}{2 \cdot 3} = \frac{6 \pm \sqrt{36 + 588}}{6} = \frac{6 \pm \sqrt{624}}{6}$$ Simplify $\sqrt{624} = \sqrt{16 \cdot 39} = 4\sqrt{39}$: $$x = \frac{6 \pm 4\sqrt{39}}{6} = 1 \pm \frac{2\sqrt{39}}{3}$$ 13. Check domain $x \geq 4$: - For $x = 1 + \frac{2\sqrt{39}}{3}$, approximate $\sqrt{39} \approx 6.245$, so $$x \approx 1 + \frac{2 \times 6.245}{3} = 1 + 4.163 = 5.163 \geq 4$$ valid. - For $x = 1 - \frac{2\sqrt{39}}{3}$, approximate $$x \approx 1 - 4.163 = -3.163 < 4$$ invalid. 14. Verify the valid solution in the original equation to avoid extraneous roots. Final answer: $$x = 1 + \frac{2\sqrt{39}}{3}$$