Solve x²+ ½x - 4½ > 0: Finding Positive Values of a Quadratic Function

To solve this problem, we will analyze the quadratic function $y = x^2 + \frac{1}{2}x - 4\frac{1}{2}$. We need to determine for which values of $x$ the function $f(x) > 0$.

• Step 1: Identify the coefficients: The function can be rewritten in standard form as $y = ax^2 + bx + c$, where $a = 1$, $b = \frac{1}{2}$, $c = -\frac{9}{2}$.
• Step 2: Use the quadratic formula to find the roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant: $\Delta = b^2 - 4ac = \left(\frac{1}{2}\right)^2 - 4(1)\left(-\frac{9}{2}\right) = \frac{1}{4} + 18 = \frac{73}{4}$. Since $\Delta > 0$, there are two real roots.
• Step 4: Find the roots: $x = \frac{-\frac{1}{2} \pm \sqrt{\frac{73}{4}}}{2} = \frac{-\frac{1}{2} \pm \frac{\sqrt{73}}{2}}{2}$ $x = \frac{-1 \pm \sqrt{73}}{4}$ These roots are $x_1 = \frac{-1 - \sqrt{73}}{4}$ and $x_2 = \frac{-1 + \sqrt{73}}{4}$.
• Step 5: Analyze the sign of the quadratic: The parabola opens upwards (as $a = 1 > 0$). It is positive outside its roots: $x < x_1$ and $x > x_2$.

Therefore, the values of $x$ that satisfy $f(x) > 0$ are $x > \frac{-1+\sqrt{73}}{4}$ or $x < \frac{-1-\sqrt{73}}{4}$.

The solution is $x > \frac{-1+\sqrt{73}}{4}$ or $x < \frac{-1-\sqrt{73}}{4}$.