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:
[beautiful math coming... please be patient] $x^2  9 = 0$  (original equation; check that $\,0\,$ is on one side of the equation) 
[beautiful math coming... please be patient] $(x+3)(x3) = 0$  (factor to get a product on the other side) 
[beautiful math coming... please be patient] $x+3 = 0\ \ \text{or}\ \ x3 = 0$  (use the zero factor law) 
[beautiful math coming... please be patient] $x=3\ \ \text{or}\ \ x = 3$  (solve the simpler equations) 
Notice that if
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$\,k\lt 0\,$,
then the equation $\,z^2 = k\,$ has no real number solutions.
For example, consider the equation
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$\,z^2 = 4\,$.
There is no real number which, when squared, gives $\,4\,$.
To use this approach, you must:
[beautiful math coming... please be patient] $x^2  9 = 0$  (original equation) 
[beautiful math coming... please be patient] $x^2 = 9$  (isolate a perfect square by adding $\,9\,$ to both sides) 
[beautiful math coming... please be patient] $x = \pm\sqrt{9}$  (check that $\,k\ge 0\,$; use the theorem) 
[beautiful math coming... please be patient] $x = \pm 3$  (rename: $\,\sqrt{9} = 3\,$) 
[beautiful math coming... please be patient] $x = 3\ \ \text{or}\ \ x = 3$  (expand the ‘plus or minus’ shorthand, if desired) 
Here are three slightly different approaches to solving the equation [beautiful math coming... please be patient] $\,16x^2  25 = 0\,$:
[beautiful math coming... please be patient] $16x^2  25 = 0$  (original equation) 
[beautiful math coming... please be patient] $(4x)^2  5^2 = 0$  (rewrite lefthand side as a difference of squares) 
[beautiful math coming... please be patient] $(4x + 5)(4x  5) = 0$  (factor the lefthand side) 
[beautiful math coming... please be patient] $4x + 5 = 0\ \ \text{or}\ \ 4x5 = 0$  (use the Zero Factor Law) 
[beautiful math coming... please be patient] $4x = 5\ \ \text{or}\ \ 4x = 5$  (solve the simpler equations) 
[beautiful math coming... please be patient] $\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$  (solve the simpler equations) 
[beautiful math coming... please be patient] $16x^2  25 = 0$  (original equation) 
[beautiful math coming... please be patient] $16x^2 = 25$  (add $\,25\,$ to both sides) 
[beautiful math coming... please be patient] $\displaystyle x^2 = \frac{25}{16}$  (divide both sides by $\,16\,$: now, $x^2$ is isolated) 
[beautiful math coming... please be patient] $\displaystyle x = \pm\sqrt{\frac{25}{16}}$  (use the theorem) 
[beautiful math coming... please be patient] $\displaystyle x = \pm\frac{5}{4}$  (rename: [beautiful math coming... please be patient] $\,\sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4}$) 
[beautiful math coming... please be patient] $\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$  (expand the ‘plus or minus’ shorthand, if desired) 
[beautiful math coming... please be patient] $16x^2  25 = 0$  (original equation) 
[beautiful math coming... please be patient] $16x^2 = 25$  (add $\,25\,$ to both sides) 
[beautiful math coming... please be patient] $(4x)^2 = 25$  (rename lefthand side as a perfect square) 
[beautiful math coming... please be patient] $4x = \pm\sqrt{25}$  (use the theorem) 
[beautiful math coming... please be patient] $4x = \pm 5$  (rename: $\,\sqrt{25} = 5\,$) 
[beautiful math coming... please be patient] $\displaystyle x = \frac{\pm 5}{4}$  (divide both sides by $\,4\,$) 
[beautiful math coming... please be patient] $\displaystyle x = \frac{5}{4}\ \ \text{or}\ \ x = \frac{5}{4}$  (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:
