1. **State the problem:** Solve the inequality $$\sqrt{2x - 4} + \sqrt{x + 4} \geq \sqrt{2x}$$. 2. **Identify domain restrictions:** - Inside each square root must be non-negative: - $2x - 4 \geq 0 \Rightarrow x \geq 2$ - $x + 4 \geq 0 \Rightarrow x \geq -4$ - $2x \geq 0 \Rightarrow x \geq 0$ The domain is the intersection: $x \geq 2$. 3. **Rewrite the inequality:** $$\sqrt{2x - 4} + \sqrt{x + 4} \geq \sqrt{2x}$$ 4. **Isolate one square root:** $$\sqrt{2x - 4} \geq \sqrt{2x} - \sqrt{x + 4}$$ 5. **Square both sides carefully:** $$2x - 4 \geq (\sqrt{2x} - \sqrt{x + 4})^2 = 2x - 2\sqrt{2x(x + 4)} + (x + 4)$$ 6. **Simplify:** $$2x - 4 \geq 2x - 2\sqrt{2x(x + 4)} + x + 4$$ Subtract $2x$ from both sides: $$-4 \geq -2\sqrt{2x(x + 4)} + x + 4$$ 7. **Rearrange:** $$-4 - x - 4 \geq -2\sqrt{2x(x + 4)}$$ $$-x - 8 \geq -2\sqrt{2x(x + 4)}$$ Multiply both sides by $-1$ (reverse inequality): $$x + 8 \leq 2\sqrt{2x(x + 4)}$$ 8. **Square both sides again:** $$(x + 8)^2 \leq 4 \cdot 2x(x + 4) = 8x(x + 4)$$ 9. **Expand and simplify:** $$(x + 8)^2 = x^2 + 16x + 64$$ $$8x(x + 4) = 8x^2 + 32x$$ Inequality: $$x^2 + 16x + 64 \leq 8x^2 + 32x$$ Bring all terms to one side: $$0 \leq 8x^2 + 32x - x^2 - 16x - 64$$ $$0 \leq 7x^2 + 16x - 64$$ 10. **Solve quadratic inequality:** $$7x^2 + 16x - 64 \geq 0$$ Find roots using quadratic formula: $$x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 7 \cdot (-64)}}{2 \cdot 7} = \frac{-16 \pm \sqrt{256 + 1792}}{14} = \frac{-16 \pm \sqrt{2048}}{14}$$ Simplify $\sqrt{2048} = \sqrt{1024 \cdot 2} = 32\sqrt{2}$: $$x = \frac{-16 \pm 32\sqrt{2}}{14} = \frac{-8 \pm 16\sqrt{2}}{7}$$ Approximate roots: - $x_1 = \frac{-8 - 16\sqrt{2}}{7} \approx -5.66$ - $x_2 = \frac{-8 + 16\sqrt{2}}{7} \approx 1.61$ 11. **Analyze inequality:** Since leading coefficient $7 > 0$, quadratic is positive outside roots. So, $$x \leq -5.66 \quad \text{or} \quad x \geq 1.61$$ 12. **Combine with domain $x \geq 2$:** Only $x \geq 2$ satisfies domain and inequality. **Final solution:** $$\boxed{x \geq 2}$$