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 $\,(3x+2)^2 = 16\,$.
To use this approach, you must:
$\,(3x+2)^2 = 16\,$  (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) 
$(3x+2+4)(3x+24) = 0$  (factor the difference of squares) 
$(3x+6)(3x2) = 0$  (simplify) 
$3x+6 = 0\ \ \text{or}\ \ 3x2 = 0$  (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) 
$(3(2)+2)^2\ \ \overset{\text{?}}{=}\ \ 16$  $(3(\frac{2}{3})+2)^2\ \ \overset{\text{?}}{=}\ \ 16$ 
$(6 + 2)^2 \ \ \overset{\text{?}}{=} \ \ 16$  $(2 + 2)^2 \ \ \overset{\text{?}}{=} \ \ 16$ 
$(4)^2 \ \ \overset{\text{?}}{=} \ \ 16$  $(4)^2 \ \ \overset{\text{?}}{=} \ \ 16$ 
$16 = 16$ Check!  $16 = 16$ Check! 
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:
$(3x+2)^2 = 16$  (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 $\,3\,$) 
For more advanced students, a graph is displayed.
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 if you prefer not to see the graph.
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.However, you can check to see if your answer is correct. 
PROBLEM TYPES:
