Question

what are the solutions to the quadratic equation $x^2 - 16 = 0$?
$\bigcirc$ $x = 2$ and $x = -2$
$\bigcirc$ $x = 4$ and $x = -4$
$\bigcirc$ $x = 8$ and $x = -8$
$\bigcirc$ $x = 16$ and $x = -16$

Explanation:

Step1: Rearrange the equation

We start with the equation \(x^{2}-16 = 0\). Add 16 to both sides of the equation to isolate the \(x^{2}\) term.
\(x^{2}=16\)

Step2: Solve for x

To solve for \(x\), we take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative solution because \((a)^{2}=(-a)^{2}\).
So, \(x=\pm\sqrt{16}\)
Since \(\sqrt{16} = 4\), we have \(x = 4\) or \(x=- 4\)

Answer:

\(x = 4\) and \(x=-4\) (corresponding to the option: \(x = 4\) and \(x=-4\))