Quadratic Inequality 313De6

1. **State the problem:** We need to find all values of $x$ such that $$k(x) = 3x^2 - 19x - 14 > 0.$$ 2. **Recall the formula and rules:** For a quadratic function $ax^2 + bx + c$, the sign of the function depends on the roots and the leading coefficient $a$. Since $a=3 > 0$, the parabola opens upward. The function is positive outside the roots and negative between them. 3. **Find the roots by solving $k(x) = 0$: $$3x^2 - 19x - 14 = 0.$$** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{19 \pm \sqrt{(-19)^2 - 4 \cdot 3 \cdot (-14)}}{2 \cdot 3}.$$ Calculate the discriminant: $$(-19)^2 - 4 \cdot 3 \cdot (-14) = 361 + 168 = 529.$$ Square root: $$\sqrt{529} = 23.$$ So the roots are: $$x = \frac{19 \pm 23}{6}.$$ 4. **Calculate each root:** $$x_1 = \frac{19 - 23}{6} = \frac{-4}{6} = -\frac{2}{3},$$ $$x_2 = \frac{19 + 23}{6} = \frac{42}{6} = 7.$$ 5. **Determine where $k(x) > 0$:** Since $a=3 > 0$, the parabola opens upward, so $k(x) > 0$ when $x < -\frac{2}{3}$ or $x > 7$. **Final answer:** $$\boxed{\left(-\infty, -\frac{2}{3}\right) \cup (7, \infty)}.$$