Quadratic Inequality 9Ac6C4

1. **State the problem:** Solve the inequality $$2x^2 + 3x - 2 < 0$$. 2. **Formula and rules:** To solve quadratic inequalities, first find the roots of the quadratic equation $$2x^2 + 3x - 2 = 0$$ by factoring or using the quadratic formula. Then determine the intervals where the quadratic expression is less than zero. 3. **Find the roots:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=2$$, $$b=3$$, and $$c=-2$$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 3^2 - 4 \times 2 \times (-2) = 9 + 16 = 25$$. Calculate the roots: $$x = \frac{-3 \pm \sqrt{25}}{2 \times 2} = \frac{-3 \pm 5}{4}$$. So, $$x_1 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$ $$x_2 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$. 4. **Determine intervals:** The roots split the number line into three intervals: $$(-\infty, -2)$$, $$(-2, \frac{1}{2})$$, and $$\left(\frac{1}{2}, \infty\right)$$. 5. **Test each interval:** Choose test points in each interval to check the sign of $$2x^2 + 3x - 2$$. - For $$x = -3$$ (in $$(-\infty, -2)$$): $$2(-3)^2 + 3(-3) - 2 = 18 - 9 - 2 = 7 > 0$$ - For $$x = 0$$ (in $$(-2, \frac{1}{2})$$): $$2(0)^2 + 3(0) - 2 = -2 < 0$$ - For $$x = 1$$ (in $$\left(\frac{1}{2}, \infty\right)$$): $$2(1)^2 + 3(1) - 2 = 2 + 3 - 2 = 3 > 0$$ 6. **Conclusion:** The quadratic expression is less than zero on the interval $$\boxed{(-2, \frac{1}{2})}$$. **Final answer:** $$x \in (-2, \frac{1}{2})$$.