Compare Square Expressions: (x-4)² vs x²+16 with x>0

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

• Step 1: Expand both expressions.
• Step 2: Subtract one expression from the other to determine the inequality.
• Step 3: Analyze and conclude based on the comparison.

Now, let's work through each step:
Step 1: Expand $(x-4)^2$:

$(x-4)^2 = (x-4)(x-4) = x^2 - 2 \cdot 4 \cdot x + 16 = x^2 - 8x + 16$

Step 2: Set up the inequality $(x-4)^2 < x^2 + 16$ and substitute the expanded form:

$x^2 - 8x + 16 < x^2 + 16$

Step 3: Simplify the inequality by subtracting $x^2$ and $16$ from both sides:

$x^2 - 8x + 16 - x^2 - 16 < 0$

$-8x < 0$

Solving $-8x < 0$ gives:

$x > 0$

This inequality holds true for all $x > 0$.

Therefore, the inequality $(x-4)^2 < x^2 + 16$ is correct for $x > 0$.

Thus, the correct symbol to fill in the blank is $\lt$.