1. Let's start by stating the problem: A quadratic inequality involves an expression of the form $ax^2 + bx + c$ where $a \neq 0$, and we want to find the values of $x$ that make the inequality true, such as $ax^2 + bx + c > 0$, $ax^2 + bx + c < 0$, $ax^2 + bx + c \geq 0$, or $ax^2 + bx + c \leq 0$. 2. The general form of a quadratic inequality is: $$ax^2 + bx + c \; \square \; 0$$ where $\square$ can be $>$, $<$, $\geq$, or $\leq$. 3. Important rules: - The quadratic expression $ax^2 + bx + c$ is a parabola when graphed. - The sign of $a$ determines if the parabola opens upwards ($a > 0$) or downwards ($a < 0$). - The roots (solutions) of the quadratic equation $ax^2 + bx + c = 0$ divide the number line into intervals. 4. To solve a quadratic inequality: - Find the roots by solving $ax^2 + bx + c = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ - These roots split the number line into intervals. - Test a point from each interval in the inequality to see if it satisfies the inequality. 5. Example: Solve $x^2 - 5x + 6 > 0$. - Find roots: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}$$ - Roots are $x=2$ and $x=3$. - Intervals: $(-\infty, 2)$, $(2, 3)$, $(3, \infty)$. - Test $x=1$ in $x^2 - 5x + 6$: $1 - 5 + 6 = 2 > 0$ (true). - Test $x=2.5$: $6.25 - 12.5 + 6 = -0.25 < 0$ (false). - Test $x=4$: $16 - 20 + 6 = 2 > 0$ (true). - So solution is $x < 2$ or $x > 3$. 6. Summary: Quadratic inequalities are solved by finding roots, testing intervals, and determining where the quadratic expression is positive or negative according to the inequality sign.