Quadratic Inequality A692A2

1. **State the problem:** Solve the inequality $$x^2 + 6x \geq -5$$. 2. **Rewrite the inequality:** Add 5 to both sides to get all terms on one side: $$x^2 + 6x + 5 \geq 0$$ 3. **Factor the quadratic:** $$x^2 + 6x + 5 = (x + 1)(x + 5)$$ 4. **Analyze the inequality:** We want to find where $$(x + 1)(x + 5) \geq 0$$ 5. **Find critical points:** The roots are at $x = -1$ and $x = -5$. 6. **Test intervals:** - For $x < -5$, pick $x = -6$: $( -6 + 1)( -6 + 5) = (-5)(-1) = 5 \geq 0$ (True) - For $-5 < x < -1$, pick $x = -3$: $( -3 + 1)( -3 + 5) = (-2)(2) = -4 < 0$ (False) - For $x > -1$, pick $x = 0$: $(0 + 1)(0 + 5) = 1 \times 5 = 5 \geq 0$ (True) 7. **Include equality:** Since the inequality is $\geq 0$, include $x = -5$ and $x = -1$. 8. **Final solution:** $$x \in (-\infty, -5] \cup [-1, \infty)$$