1. State the problem: Solve the inequality $$\sqrt{5 - x} \geq x - 3$$. 2. Consider the domain: The expression inside the square root must be non-negative: $$5 - x \geq 0 \implies x \leq 5.$$ 3. Also, since the square root is always non-negative, the right side must be less or equal to the left side. This means $$x - 3 \leq \sqrt{5 - x}$$. 4. To solve, isolate terms by squaring both sides (remember to check for extraneous solutions): $$\sqrt{5 - x} \geq x - 3 \implies (\sqrt{5 - x})^2 \geq (x-3)^2$$ $$5 - x \geq (x - 3)^2 = x^2 - 6x + 9.$$ 5. Rearrangement: $$5 - x \geq x^2 - 6x + 9 \implies 0 \geq x^2 - 5x + 4.$$ 6. Rewrite the inequality: $$x^2 - 5x + 4 \leq 0.$$ 7. Factor the quadratic: $$x^2 - 5x + 4 = (x - 4)(x - 1).$$ 8. The inequality becomes: $$(x - 4)(x - 1) \leq 0.$$ 9. The product of two terms is less than or equal to zero when the terms have opposite signs or one is zero, which corresponds to: $$1 \leq x \leq 4.$$ 10. Combine this with the domain constraint: $$x \leq 5$$ from step 2, so the feasible range is $$1 \leq x \leq 4.$$ 11. Verify boundary points by substituting back into the original inequality: - For $$x=1$$, left side: $$\sqrt{5-1} = \sqrt{4} = 2$$; right side: $$1-3 = -2$$; inequality holds. - For $$x=4$$, left: $$\sqrt{5-4} = \sqrt{1} = 1$$; right: $$4-3=1$$; inequality holds. 12. Check for any extraneous solution outside $$[1,4]$$ in domain: - For $$x=0$$, left: $$\sqrt{5-0} = \sqrt{5} \approx 2.236$$; right: $$0-3 = -3$$; holds but $$0$$ is outside interval from step 9. However, check if inequality holds for $$x=0$$: $$\sqrt{5-0} = 2.236 \geq -3$$ is true but since $$x=0$$ makes expression positive, check domain both satisfied; however, it conflicts with step 9, implying we must respect the quadratic condition for the inequality after squaring. 13. Test a point outside $$[1,4]$$ but inside domain: - $$x=0$$ satisfies initial inequality, so include it. - However, from step 9, we concluded $$1 \leq x \leq 4$$ from squaring; the difference arises because squaring can produce extraneous results. 14. To incorporate both findings, check the original inequality over domain $$x \leq 5$$ by testing points: - At $$x=0$$, inequality holds. - At $$x=5$$, left: $$\sqrt{0} = 0$$, right: $$5 - 3 = 2$$, so fails. - At $$x=3$$, left: $$\sqrt{5-3} = \sqrt{2} \approx 1.414$$, right: $$3-3=0$$, holds. 15. By analysis, the solution is all $$x$$ such that $$x \leq 4$$ and $$x \geq -1$$? Check for $$x=-1$$: - $$\sqrt{5 - (-1)} = \sqrt{6} \approx 2.449$$, right: $$-1-3 = -4$$, holds. 16. Checking whether negative large $$x$$ satisfies: - At $$x=-10$$, left: $$\sqrt{5 - (-10)} = \sqrt{15} \approx 3.87$$, right: $$-10 - 3 = -13$$, true. 17. We must pay attention to the original inequality form involving the square root. Split into cases: Case 1: $$x - 3 \leq 0$$ (i.e. $$x \leq 3$$) Then $$\sqrt{5 - x} \geq x - 3$$ always true because left >= 0, right <= 0. Case 2: $$x - 3 > 0$$ (i.e. $$x > 3$$) Then square both sides: $$5 - x \geq (x - 3)^2\Rightarrow x^2 - 5x +4 \leq 0$$ which reduces to $$1 \leq x \leq 4$$. 18. Combine both cases: - For $$x \leq 3$$, all $$x$$ in domain $$x \leq 5$$ work where $$x \leq 5$$ and $$5 - x \geq 0$$ implies $$x \leq 5$$. - For $$3 < x \leq 5$$, quadratic condition requires $$1 \leq x \leq 4$$, so intersection is $$3 < x \leq 4$$. 19. Final solution is all $$x$$ satisfying: $$x \leq 3$$ (from case 1) OR $$3 < x \leq 4$$ (from case 2) which simplifies to $$x \leq 4$$ and must also satisfy the domain $$x \leq 5$$. 20. Hence the solution set is: $$\boxed{(-\infty, 4]}.$$