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

Question

Look at the function below:

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

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

f(x) > 0

Step-by-Step Solution

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

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

Step 2: Solve the equation above for x x :

4x+22=04x=22x=224=5.5 4x + 22 = 0 \\ 4x = -22 \\ x = -\frac{22}{4} = -5.5

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

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

Therefore, the function is positive (f(x)>0) (f(x) > 0) when x5.5 x \neq -5.5 .

Thus, the solution to the problem is: x512 x \neq -5\frac{1}{2} .

Answer

x512 x\ne-5\frac{1}{2}