1. **Evaluate** $|3^2 - 5^2|$. First, calculate the squares: $$3^2 = 9, \quad 5^2 = 25$$ Then subtract: $$9 - 25 = -16$$ Take the absolute value: $$| -16 | = 16$$ 2. **Solve the inequality** $\sqrt{2x+3} > x$. - The expression under the square root must be non-negative: $$2x + 3 \geq 0 \implies x \geq -\frac{3}{2}$$ - Square both sides to remove the square root (noting to check for extraneous solutions): $$\sqrt{2x+3} > x \implies 2x + 3 > x^2$$ Rewrite as: $$x^2 - 2x - 3 < 0$$ Factor the quadratic: $$x^2 - 2x - 3 = (x - 3)(x + 1)$$ The inequality is: $$(x - 3)(x + 1) < 0$$ This holds when $x$ is between the roots: $$-1 < x < 3$$ - Combine with domain $x \geq -\frac{3}{2}$: Since $-\frac{3}{2} = -1.5 < -1$, the domain restricts to $x \geq -1.5$, but the inequality requires $x > -1$, so the solution is: $$-1 < x < 3$$ - Check endpoints and extraneous solutions: At $x = -1$, $\sqrt{2(-1)+3} = \sqrt{1} = 1$ and $x = -1$, so $1 > -1$ true. At $x = 3$, $\sqrt{2(3)+3} = \sqrt{9} = 3$ and $x=3$, so $3 > 3$ false. So solution is: $$-1 < x < 3$$ 3. **Solve the inequality** $|2x^2 - 13| < 5$. Rewrite as: $$-5 < 2x^2 - 13 < 5$$ Add 13 to all parts: $$8 < 2x^2 < 18$$ Divide all parts by 2: $$4 < x^2 < 9$$ Take square roots: $$2 < |x| < 3$$ This means: $$-3 < x < -2 \quad \text{or} \quad 2 < x < 3$$