Solve (2x-16)² > 0: Finding Values Where Function is Positive

Question

Look at the function below:

y=(2x16)2 y=\left(2x-16\right)^2

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

f(x) > 0

Step-by-Step Solution

To determine when the function y=(2x16)2 y = (2x - 16)^2 is greater than zero, we observe the following:

  • The function's expression, (2x16)2(2x - 16)^2, is a square and thus always non-negative (0 \geq 0 ).
  • The expression will equate to zero when the inside term is zero: 2x16=02x - 16 = 0.
  • Solve the equation 2x16=02x - 16 = 0 to find x=8x = 8.
  • Therefore, (2x16)2=0(2x - 16)^2 = 0 only at x=8x = 8.
  • For y>0 y > 0 , (2x16)2(2x - 16)^2 must be greater than zero, which occurs for all x x except x=8 x = 8.

The solution is x8 x \ne 8 , which means y y is positive for all x x except x=8 x = 8 .

Answer

x8 x\ne8