# Solving More Complicated Equations Involving Perfect Squares

Need some simpler practice first?

Here, you will solve more complicated equations involving perfect squares. As in the previous section, there are two basic approaches you can use. They're both discussed thoroughly on this page.

The two approaches are illustrated next, by solving the equation $\,(3x+2)^2 = 16\,.$

## Approach #1: Factor and use the Zero Factor Law

To use this approach, you must:

• Get $\,0\,$ on one side of the equation.
• Factor to get a product on the other side of the equation.
• Use the Zero Factor Law: For all real numbers $\,a\,$ and $\,b\,,$ $\,ab = 0\,$ is equivalent to $\,(a = 0\ \ \text{or}\ \ b = 0)\,.$

To use this approach, you would write the following list of equivalent sentences:

 $\,(3x+2)^2 = 16\,$ original equation $\,(3x+2)^2 -16 = 0\,$ need $\,0\,$ on one side; subtract $\,16\,$ from both sides $(3x+2)^2 - 4^2 = 0$ rewrite, so it's clear you have a difference of squares $(3x+2+4)(3x+2-4) = 0$ factor the difference of squares $(3x+6)(3x-2) = 0$ simplify $3x+6 = 0\ \ \text{or}\ \ 3x-2 = 0$ use the Zero Factor Law $3x = -6\ \ \text{or}\ \ 3x = 2$ solve the simpler equations $\displaystyle x = -2\ \ \text{or}\ \ x = \frac{2}{3}$ solve the simpler equations

It's a good idea to check:

$$\begin{gather} \cssId{s33}{(3(-2)+2)^2\ \ \overset{\text{?}}{=}\ \ 16}\cr \cssId{s34}{(-6 + 2)^2 \ \ \overset{\text{?}}{=} \ \ 16}\cr \cssId{s35}{(-4)^2 \ \ \overset{\text{?}}{=} \ \ 16}\cr \cssId{s36}{16 = 16\ \ \ \ \text{Check!}}\cr\cr \end{gather}$$ $$\begin{gather} \cssId{s37}{(3(\frac{2}{3})+2)^2\ \ \overset{\text{?}}{=}\ \ 16}\cr \cssId{s38}{(2 + 2)^2 \ \ \overset{\text{?}}{=} \ \ 16}\cr \cssId{s39}{(4)^2 \ \ \overset{\text{?}}{=} \ \ 16}\cr \cssId{s40}{16 = 16\ \ \ \ \text{Check!}} \end{gather}$$

## Approach #2:Use the Following Theorem

THEOREM solving equations involving perfect squares
For all real numbers $\,z\$ and for $\,k\ge 0\,$: $$\cssId{s46}{z^2 = k\ \ \ \text{is equivalent to}\ \ \ z=\pm\sqrt{k}}$$

The basic idea is that you're (correctly!) ‘undoing’ a square with the square root. Notice that if $\,k\lt 0\,,$ then the equation $\,z^2 = k\,$ has no real number solutions. For example, consider the equation $\,z^2 = -4\,.$ There is no real number which, when squared, gives $\,-4\,.$

To use this approach, you must:

• Isolate a perfect square on one side of the equation.
• Check that you have a nonnegative number on the other side.
• Use the theorem.

To use this approach, you would write the following list of equivalent sentences:

 $(3x+2)^2 = 16$ original equation $3x + 2 = \pm\sqrt{16}$ check that $\,k\ge 0\,$; use the theorem $3x + 2 = \pm 4$ simplify: $\,\sqrt{16} = 4\,$ $3x + 2 = 4\ \ \text{or}\ \ 3x + 2 = -4$ expand the ‘plus or minus’ shorthand) $3x = 2\ \ \text{or}\ \ 3x = -6$ subtract $\,2\,$ from both sides of both equations $\displaystyle x = \frac{2}{3}\ \ \text{or}\ \ x = -2$ divide both sides of both equations by $\,3\,$

## Concept Practice

For more advanced students, a graph is available. For example, the equation $\,(3x+2)^2 = 16\,$ is optionally accompanied by the graph of $\,y = (3x+2)^2\,$ (the left side of the equation, dashed green) and the graph of $\,y = 16\,$ (the right side of the equation, solid purple). In this example, you are finding the values of $\,x\,$ where the green graph intersects the purple graph.

Click the ‘Show/Hide Graph’ button to toggle the graph.

Solve: