Inequality Solution 918C5D

1. **State the problem:** Solve the inequality $$x^2 - \frac{4}{x} + 3 > 0$$ and express the solution as a union of intervals and on a number line. 2. **Rewrite the inequality:** To handle the fraction, multiply both sides by $x$ (noting $x \neq 0$) and consider cases based on the sign of $x$. 3. **Case 1: $x > 0$** Multiply both sides by $x > 0$ (inequality direction stays the same): $$x \left(x^2 - \frac{4}{x} + 3\right) > 0 \implies x^3 - 4 + 3x > 0$$ Simplify: $$x^3 + 3x - 4 > 0$$ 4. **Solve $x^3 + 3x - 4 = 0$:** Try rational roots: $x=1$: $$1^3 + 3(1) - 4 = 1 + 3 - 4 = 0$$ So $x=1$ is a root. Factor: $$x^3 + 3x - 4 = (x - 1)(x^2 + x + 4)$$ Since $x^2 + x + 4$ has discriminant $1 - 16 = -15 < 0$, no real roots. 5. **Sign analysis for $x^3 + 3x - 4$:** - For $x > 1$, $(x-1) > 0$ and $x^2 + x + 4 > 0$, so product $> 0$. - For $0 < x < 1$, $(x-1) < 0$ and $x^2 + x + 4 > 0$, so product $< 0$. 6. **Case 2: $x < 0$** Multiply both sides by $x < 0$ (inequality direction reverses): $$x \left(x^2 - \frac{4}{x} + 3\right) > 0 \implies x^3 - 4 + 3x > 0$$ Reversing inequality: $$x^3 + 3x - 4 < 0$$ 7. **Sign analysis for $x^3 + 3x - 4 < 0$ when $x < 0$:** Check values: - At $x = -1$, $(-1)^3 + 3(-1) - 4 = -1 - 3 - 4 = -8 < 0$ - At $x = 0$, $0 + 0 - 4 = -4 < 0$ Since the cubic is continuous and no roots less than zero, it is negative for all $x < 0$. 8. **Exclude $x=0$** because of division by zero. 9. **Combine results:** - For $x < 0$, inequality holds. - For $0 < x < 1$, inequality does not hold. - For $x > 1$, inequality holds. 10. **Final solution:** $$(-\infty, 0) \cup (1, \infty)$$ 11. **Number line:** - Shade all values less than $0$. - Leave $(0,1)$ unshaded. - Shade all values greater than $1$. - Mark points $0$ and $1$ as open circles (excluded). **Answer:** The solution to $$x^2 - \frac{4}{x} + 3 > 0$$ is $$(-\infty, 0) \cup (1, \infty)$$.