Solve (4x+22)² > 0: Finding Values of x in a Perfect Square Inequality

Look at the function below:

$y=\left(4x+22\right)^2$

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

$f(x) > 0$

To solve this problem, we observe that the function given is $y = (4x + 22)^2$.

Step 1: We set the expression inside the square equal to zero and solve for $x$.
$4x + 22 = 0$

Step 2: Solve the equation above for $x$:

This calculation reveals that $x = -5.5$ is the only point where $(4x + 22)^2 = 0$.

Step 3: Outside of this specific $x$, the squared term $(4x+22)^2$ is positive for all other values of $x$.

Therefore, the function is positive $(f(x) > 0)$ when $x \neq -5.5$.

Thus, the solution to the problem is: $x \neq -5\frac{1}{2}$.