Solve (2x-16)² < 0: Finding Values of x in a Quadratic Inequality

Look at the function below:

$y=\left(2x-16\right)^2$

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

$f(x) < 0$

To solve this problem, we must determine when the expression $(2x-16)^2$ is less than zero.

First, consider the expression $(2x-16)^2$.

• Recognize that squaring any real number results in a non-negative number. Thus, $(2x-16)^2$ is always non-negative.
• Specifically, for any expression squared, $(a)^2 \geq 0$ for all real numbers $a$.
• Therefore, $(2x-16)^2$ cannot be less than zero for any value of $x$.

Since a square of any real function is always zero or positive, there are no real values of $x$ for which $(2x-16)^2$ is negative.

Therefore, the conclusion is that there are no values of $x$ that make $(2x-16)^2 < 0$.

The correct answer is: No $x$.