Solve (x-6)(x+6) > 0: Finding Positive Values of a Quadratic Function

Look at the following function:

$y=\left(x-6\right)\left(x+6\right)$

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

$f(x) > 0$

The function $y = (x-6)(x+6)$ can be rewritten as $y = x^2 - 36$. This is a quadratic function, and we need to find where it is positive: $x^2 - 36 > 0$.

First, identify the roots of the quadratic equation $x^2 - 36 = 0$. Solving for $x$, we get:

• $x^2 = 36$
• $x = \pm 6$

Thus, the roots are $x = 6$ and $x = -6$.

Next, examine the intervals determined by these roots: $(-\infty, -6)$, $(-6, 6)$, $(6, \infty)$.

For each interval, we check the sign of $x^2 - 36$ to determine where the expression is positive.

1. **Interval $(-\infty, -6)$:** Choose $x = -7$:
$(x-6)(x+6) = (-7-6)(-7+6) = (-13)(-1) = 13$, which is positive.

2. **Interval $(-6, 6)$:** Choose $x = 0$:
$(x-6)(x+6) = (0-6)(0+6) = (-6)(6) = -36$, which is negative.

3. **Interval $(6, \infty)$:** Choose $x = 7$:
$(x-6)(x+6) = (7-6)(7+6) = (1)(13) = 13$, which is positive.

Therefore, the quadratic $x^2 - 36 > 0$ in the intervals $(-\infty, -6)$ and $(6, \infty)$. The function is positive on these intervals.

Since the solution matches choice id="4", the values of $x$ for which $f(x) > 0$ are:
$x > 6$ or $x < -6$.

Thus, the solution to the problem is $x > 6$ or $x < -6$.