In this section, we begin to solve algebraic equations containing the square of our variable x,

$$x^2 + a = b,$$

where a and b are constants. Solving these involves taking a square root, the inverse operation of squaring. Remember that

$$\sqrt{x^2} = (\sqrt{x})^2 = x$$

Notice that the squaring and square-root operations (functions, really) undo one another. Applying them in sequence is like doing nothing at all to the variable x. They allow us to "liberate" a variable from a root or square.

Here's an example of this class of problem:

$$x^2 - 4 = 12$$

As usual, the first step to solving such a problem is to do the easy stuff first, namely, move the $-4$ to the right by adding it to both sides:

$$ \begin{align} &x^2 - 4 = 12 \\ &\underline{\phantom{00} + 4 \phantom{0} + 4} \\ &\phantom{0000}x^2 = 16 \end{align}$$

Now to find x, we take the square-root of both sides:

$$ \begin{align} \sqrt{x^2} &= ±\sqrt{16}\\ x &= ±4 \end{align}$$

Notice that we get two solutions each time we take a square root because $(4)^2 = 16 \; \color{#E90F89}{\text{ or }} \; (-4)^2 = 16.$ It's important to remember that.