Solve the Quadratic Inequality: Understanding y = -2x² + 8x - 6

Look at the following function:

$y=-2x^2+8x-6$

Determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, we'll follow these steps:

• Step 1: Determine the roots of the quadratic function.
• Step 2: Split the number line based on these roots.
• Step 3: Test intervals to check where the function is greater than zero.
• Step 4: Identify the correct interval corresponding to the solution.

Now, let's work through each step:

Step 1: Identify the roots of the quadratic equation $-2x^2 + 8x - 6 = 0$. Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -2$, $b = 8$, and $c = -6$.

Calculate the discriminant: $b^2 - 4ac = 8^2 - 4(-2)(-6) = 64 - 48 = 16$.

The roots are: $x = \frac{-8 \pm \sqrt{16}}{-4} = \frac{-8 \pm 4}{-4}$.

Thus, $x_1 = \frac{-8 + 4}{-4} = 1$ and $x_2 = \frac{-8 - 4}{-4} = 3$.

Step 2: The roots 1 and 3 split the number line into intervals: $(-\infty, 1)$, $(1, 3)$, $(3, \infty)$.

Step 3: Test a sample point from each interval:

• For $x = 0$ in $(-\infty, 1)$: $f(0) = -6$, which is not greater than 0.
• For $x = 2$ in $(1, 3)$: $f(2) = -2(2)^2 + 8(2) - 6 = 8 - 6 = 2$, which is greater than 0.
• For $x = 4$ in $(3, \infty)$: $f(4) = -14$, which is not greater than 0.

Step 4: We conclude that the function $-2x^2 + 8x - 6$ is greater than 0 only in the interval $(1, 3)$.

Therefore, the solution to the problem is $1 < x < 3$.