Solve the Equation: (x+4)(x-4) + (x+2)(x-2) = 0 Using Difference of Squares

First, recognize that both expressions are differences of squares.

For the first term: $(x+4)(x-4)$ can be expanded to $x^2 - 16$.

For the second term: $(x+2)(x-2)$ can be expanded to $x^2 - 4$.

Combine these equations:

$x^2 - 16 + x^2 - 4 = 0$

Simplify to:

$2x^2 - 20 = 0$

Divide the whole equation by 2:

$x^2 - 10 = 0$

Add 10 to both sides:

$x^2 = 10$

Take the square root of both sides:

$x = ±\sqrt{10}$