Solve y = x² + 10x + 16: Finding Values Where Function is Positive

Look at the function below:

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

Then determine for which values of $x$ the following is true:

$f(x) > 0$

We begin by solving for the roots of the equation $y = x^2 + 10x + 16$ by setting $f(x) = 0$.

This yields the equation $x^2 + 10x + 16 = 0$.

We use the quadratic formula $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$ to find the roots.

Here, $a = 1$, $b = 10$, and $c = 16$.

First, calculate the discriminant: $b^2 - 4ac = 10^2 - 4 \times 1 \times 16 = 100 - 64 = 36$.

The roots are then $x = \frac{{-10 \pm \sqrt{36}}}{2} = \frac{{-10 \pm 6}}{2}$.

This gives the roots $x_1 = \frac{{-10 + 6}}{2} = -2$ and $x_2 = \frac{{-10 - 6}}{2} = -8$.

The roots divide the real number line into three intervals: $x < -8$, $-8 < x < -2$, and $x > -2$.

We need to determine where the function is greater than zero, $f(x) > 0$:

• For $x < -8$, choose a test point such as $x = -9$. Calculating $f(-9) = (-9)^2 + 10(-9) + 16 = 81 - 90 + 16 = 7$, which is positive.
• For $-8 < x < -2$, choose a test point such as $x = -5$. Calculating $f(-5) = (-5)^2 + 10(-5) + 16 = 25 - 50 + 16 = -9$, which is negative.
• For $x > -2$, choose a test point such as $x = 0$. Calculating $f(0) = (0)^2 + 10(0) + 16 = 16$, which is positive.

Therefore, the solution set where $f(x) > 0$ is $x < -8$ or $x > -2$.

Upon reviewing the provided choices, the correct answer is: $x > -2$ or $x < -8$.