Solve the Quadratic Inequality: -x² + 3x + 4 > 0

To solve this quadratic inequality, follow these steps:

• Step 1: Solve the corresponding equation $-x^2 + 3x + 4 = 0$ to find critical points.
• Step 2: Test intervals between critical points to determine where the inequality holds.

Step 1: Solve the equation. The given quadratic is $-x^2 + 3x + 4 = 0$. Let's rewrite it as $x^2 - 3x - 4 = 0$ by multiplying through by $-1$.

We use the quadratic formula where $a = 1$, $b = -3$, and $c = -4$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)}$

$x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}$

The solutions to this equation are:

$x_1 = \frac{3 + 5}{2} = 4$ and $x_2 = \frac{3 - 5}{2} = -1$

Step 2: Determine where the expression is positive by checking intervals:

• Interval $(-\infty, -1)$: Choose $x = -2$. Calculating the expression: $-(-2)^2 + 3(-2) + 4 = -4 - 6 + 4 = -6$ (negative).
• Interval $(-1, 4)$: Choose $x = 0$. Calculating: $-0^2 + 3(0) + 4 = 4$ (positive).
• Interval $(4, \infty)$: Choose $x = 5$. Calculating: $-5^2 + 3(5) + 4 = -25 + 15 + 4 = -6$ (negative).

The quadratic expression $-x^2 + 3x + 4$ is positive in the interval $(-1, 4)$. Hence, for the inequality $-x^2 + 3x + 4 > 0$, we have:

The solution to the inequality is $-1 < x < 4$.