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 quadratic equation \(x^{2}-16 = 0\). First, we can rewrite it as \(x^{2}=16\) by adding 16 to both sides of the equation.

Step2: Solve for x

To solve for \(x\), we take the square root of both sides. Remember that if \(x^{2}=a\) (where \(a\geq0\)), then \(x=\pm\sqrt{a}\). So for \(x^{2} = 16\), we have \(x=\pm\sqrt{16}\). Since \(\sqrt{16} = 4\), this means \(x = 4\) or \(x=- 4\).

Answer:

B. \(x = 4\) and \(x=-4\)