Question

g the square
he square root.
$x^2 - 49 = 0$
$x^2 - 121 = 0$

Explanation:

Step1: Isolate the \(x^2\) term

We start with the equation \(x^2 - 121 = 0\). To isolate \(x^2\), we add 121 to both sides of the equation.
\[
x^2 - 121 + 121 = 0 + 121
\]
Simplifying both sides, we get:
\[
x^2 = 121
\]

Step2: Take the square root of both sides

Now that we have \(x^2 = 121\), 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 \((a)^2 = (-a)^2\).
\[
\sqrt{x^2} = \pm\sqrt{121}
\]
Simplifying the square roots, we know that \(\sqrt{x^2}=|x|\), but since we are looking for all real solutions, we can write \(x = \pm\sqrt{121}\). And \(\sqrt{121} = 11\), so:
\[
x = \pm 11
\]

Answer:

\(x = \pm 11\)