Solve (4x+22)² < 0: Finding Values for Negative Quadratic Output

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

• Step 1: Analyze the given function.
• Step 2: Determine the nature of $(4x + 22)^2$.
• Step 3: Conclude whether it can ever be negative.

Step 1: We are given the function $y = (4x + 22)^2$. This is a quadratic function expressed as a square of a linear term.

Step 2: Consider the expression $(4x + 22)$. Whatever value this linear expression takes, its square, $(4x + 22)^2$, will always be non-negative. This is because the square of a real number is never negative.

Step 3: To find when $(4x + 22)^2 < 0$, we realize that since squares are non-negative, they cannot actually be negative. Thus, $(4x + 22)^2 \geq 0$ for all values of $x$, and can never be less than zero.

Therefore, no value of $x$ will make $f(x) < 0$.

The conclusion is that there is no value of $x$ for which $f(x) < 0$.