$$ x^{2}-5 \geqslant 3 $$ 2) $$ 4 \sqrt{x}

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$$ x^{2}-5 \geqslant 3 $$ 2) $$ 4 \sqrt{x}<100 $$ 3) $$ 4 x^{3}+11<43 $$ 4) $$ \frac{1}{x}+9 \geqslant-1 $$

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Certainly! Let's solve each of the given inequalities step by step. --- ### 1) $$ x^{2} - 5 \geqslant 3 $$ **Solution:** 1. **Simplify the inequality:** $$ x^{2} - 5 \geq 3 \implies x^{2} \geq 8 $$ 2. **Solve for $$x$$:** $$ x^{2} \geq 8 \implies x \leq -\sqrt{8} \quad \text{or} \quad x \geq \sqrt{8} $$ $$ \sqrt{8} = 2\sqrt{2} \approx 2.828 $$ **Solution Set:** $$ x \leq -2\sqrt{2} \quad \text{or} \quad x \geq 2\sqrt{2} $$ **Interval Notation:** $$ (-\infty, -2\sqrt{2}] \cup [2\sqrt{2}, \infty) $$ --- ### 2) $$ 4 \sqrt{x} < 100 $$ **Solution:** 1. **Isolate the square root:** $$ 4 \sqrt{x} < 100 \implies \sqrt{x} < 25 $$ 2. **Square both sides (since both sides are non-negative):** $$ \sqrt{x} < 25 \implies x < 625 $$ 3. **Consider the domain of the square root:** $$ x \geq 0 $$ **Solution Set:** $$ 0 \leq x < 625 $$ **Interval Notation:** $$ [0, 625) $$ --- ### 3) $$ 4x^{3} + 11 < 43 $$ **Solution:** 1. **Isolate the cubic term:** $$ 4x^{3} + 11 < 43 \implies 4x^{3} < 32 \implies x^{3} < 8 $$ 2. **Take the cube root of both sides:** $$ x^{3} < 8 \implies x < \sqrt[3]{8} \implies x < 2 $$ **Solution Set:** $$ x < 2 $$ **Interval Notation:** $$ (-\infty, 2) $$ --- ### 4) $$ \frac{1}{x} + 9 \geqslant -1 $$ **Solution:** 1. **Isolate the reciprocal term:** $$ \frac{1}{x} + 9 \geq -1 \implies \frac{1}{x} \geq -10 $$ 2. **Solve the inequality $$ \frac{1}{x} \geq -10 $$ by considering cases based on the sign of $$x$$:** - **Case 1: $$ x > 0 $$** $$ \frac{1}{x} \geq -10 $$ Since $$ \frac{1}{x} $$ is positive and $$-10$$ is negative, this inequality holds for all $$ x > 0 $$. - **Case 2: $$ x < 0 $$** $$ \frac{1}{x} \geq -10 $$ Multiply both sides by $$x$$ (note that $$x$$ is negative, so the inequality sign reverses): $$ 1 \leq -10x \implies x \leq -\frac{1}{10} $$ - **Case 3: $$ x = 0 $$** $$ \frac{1}{x} \text{ is undefined} $$ 3. **Combine the cases:** $$ x \leq -\frac{1}{10} \quad \text{or} \quad x > 0 $$ **Solution Set:** $$ x \leq -\frac{1}{10} \quad \text{or} \quad x > 0 $$ **Interval Notation:** $$ \left( -\infty, -\frac{1}{10} \right] \cup (0, \infty) $$ --- ### **Summary of Solutions:** 1. **$$ x^{2} - 5 \geqslant 3 $$** $$ x \leq -2\sqrt{2} \quad \text{or} \quad x \geq 2\sqrt{2} $$ **Interval Notation:** $$(-\infty, -2\sqrt{2}] \cup [2\sqrt{2}, \infty)$$ 2. **$$ 4 \sqrt{x} < 100 $$** $$ 0 \leq x < 625 $$ **Interval Notation:** $$[0, 625)$$ 3. **$$ 4x^{3} + 11 < 43 $$** $$ x < 2 $$ **Interval Notation:** $$(-\infty, 2)$$ 4. **$$ \frac{1}{x} + 9 \geqslant -1 $$** $$ x \leq -\frac{1}{10} \quad \text{or} \quad x > 0 $$ **Interval Notation:** $$\left( -\infty, -\frac{1}{10} \right] \cup (0, \infty)$$ --- If you have any further questions or need more detailed explanations, feel free to ask! 1. $$ x \leq -2\sqrt{2} $$ or $$ x \geq 2\sqrt{2} $$ 2. $$ 0 \leq x < 625 $$ 3. $$ x < 2 $$ 4. $$ x \leq -\frac{1}{10} $$ or $$ x > 0 $$

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