Solve the Quadratic Inequality: When is -x²+10x-16 Greater Than Zero?

Look at the following function:

$y=-x^2+10x-16$

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

$f(x) > 0$

To solve the problem of identifying where the function $y = -x^2 + 10x - 16$ is greater than zero, follow these steps:

• Step 1: Calculate the roots of the quadratic equation. The roots are found using the quadratic formula:

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

Given: $a = -1$, $b = 10$, $c = -16$.

The discriminant is: $b^2 - 4ac = 10^2 - 4(-1)(-16) = 100 - 64 = 36$

Calculate the roots:

$x = \frac{-10 \pm \sqrt{36}}{2(-1)} = \frac{-10 \pm 6}{-2}$

Thus, the roots are:

• $x_1 = \frac{-10 + 6}{-2} = 2$
• $x_2 = \frac{-10 - 6}{-2} = 8$

Step 2: Determine where the function is positive. Since the parabola opens downward ($a = -1 < 0$), it is above the x-axis between the roots.

• Check intervals: $(-\infty, 2)$, $(2, 8)$, and $(8, \infty)$.

Test a point in the interval $(2, 8)$, for example, $x = 5$:

$f(5) = -5^2 + 10 \times 5 - 16 = -25 + 50 - 16 = 9 > 0$

Thus, the function is positive for $2 < x < 8$.

Conclusion: The solution to $f(x) > 0$ is $2 < x < 8$.