Here, you will solve simple equations involving perfect squares.
There are two basic approaches you can use.
They're both discussed thoroughly on this page.
Both approaches are illustrated next, using the equation $\,x^2  9 = 0\,$.
To use this approach, you must:
$x^2  9 = 0$  (original equation; check that $\,0\,$ is on one side of the equation) 
$(x+3)(x3) = 0$  (factor to get a product on the other side) 
$x+3 = 0\ \ \text{or}\ \ x3 = 0$  (use the zero factor law) 
$x=3\ \ \text{or}\ \ x = 3$  (solve the simpler equations) 
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:
$x^2  9 = 0$  (original equation) 
$x^2 = 9$  (isolate a perfect square by adding $\,9\,$ to both sides) 
$x = \pm\sqrt{9}$  (check that $\,k\ge 0\,$; use the theorem) 
$x = \pm 3$  (rename: $\,\sqrt{9} = 3\,$) 
$x = 3\ \ \text{or}\ \ x = 3$  (expand the ‘plus or minus’ shorthand, if desired) 
Here are three slightly different approaches to solving the equation $\,16x^2  25 = 0\,$:
$16x^2  25 = 0$  (original equation) 
$(4x)^2  5^2 = 0$  (rewrite lefthand side as a difference of squares) 
$(4x + 5)(4x  5) = 0$  (factor the lefthand side) 
$4x + 5 = 0\ \ \text{or}\ \ 4x5 = 0$  (use the Zero Factor Law) 
$4x = 5\ \ \text{or}\ \ 4x = 5$  (solve the simpler equations) 
$\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$  (solve the simpler equations) 
$16x^2  25 = 0$  (original equation) 
$16x^2 = 25$  (add $\,25\,$ to both sides) 
$\displaystyle x^2 = \frac{25}{16}$  (divide both sides by $\,16\,$: now, $x^2$ is isolated) 
$\displaystyle x = \pm\sqrt{\frac{25}{16}}$  (use the theorem) 
$\displaystyle x = \pm\frac{5}{4}$  (rename: $\,\sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4}$) 
$\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$  (expand the ‘plus or minus’ shorthand, if desired) 
$16x^2  25 = 0$  (original equation) 
$16x^2 = 25$  (add $\,25\,$ to both sides) 
$(4x)^2 = 25$  (rename lefthand side as a perfect square) 
$4x = \pm\sqrt{25}$  (use the theorem) 
$4x = \pm 5$  (rename: $\,\sqrt{25} = 5\,$) 
$\displaystyle x = \frac{\pm 5}{4}$  (divide both sides by $\,4\,$) 
$\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$  (expand the ‘plus or minus’ shorthand, if desired) 
For more advanced students, a graph is displayed.
For example, the equation $\,x^2  9 = 0\,$
is optionally accompanied by the
graph of $\,y = x^2  9\,$ (the left side of the equation, dashed green)
and the graph of
$\,y = 0\,$ (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:
